Factoring Trinomials

πŸ†Practice factoring trinomials

I present to you the following trinomial

ax2+bx+cax^2+bx+c

Regardless of whether the coefficients of the terms are positive or negative, as long as they appear in the style of a trinomial, the exercise will be called "trinomial".

The factorization will look like this:

(x+solutionΒ one)(x+solutionΒ two) (x+solution \space one)(x+solution\space two)
or with subtractions, depending on the solutions.

Start practice

Test yourself on factoring trinomials!

einstein

\( x^2+6x+9=0 \)

What is the value of X?

Practice more now

The first way to factor a trinomial

We will look for two numbers whose product is aΓ—c a\times c and whose sum is bb
We will ask ourselves: which number multiplied by which other will give us aΓ—c a\times c or ​​c​​c (if aa equals 11).
and what plus what would add up to bb.

In fact, we need to find a pair of numbers that meet these two conditions at the same time.

We can plot it as follows:

We can plot it as follows


The second way to factor a trinomial - quadratic formula

x=βˆ’bΒ±b2βˆ’4ac2ax = {-b \pm \sqrt{b^2-4ac} \over 2a}

aa Β Β The coefficient of the first term
bbΒ The coefficient of the second term
ccΒ The constant term

In the first step, we will use only addition to find the first solution, and then, we will use only subtraction to find the second.
Again, the factorization will look as follows:
(x+solutionΒ one)(x+solutionΒ two) (x+solution \space one)(x+solution\space two)
or with subtractions, depending on the solutions.


Join Over 30,000 Students Excelling in Math!
Endless Practice, Expert Guidance - Elevate Your Math Skills Today
Test your knowledge

What is a trinomial?

ax2+bx+cax^2+bx+c

The trinomial represents an expression in which xx is squared, preceded by a coefficient (which can be positive or negative), but it must not be 00 (sometimes the coefficient is equal to 11 and therefore we will not see the aa), to this term may be added or subtracted some other bxbx when bb represents the coefficient (under the same conditions as aa) and the independent variable (number cc) is added or subtracted.
Regardless of whether the coefficients of the terms are positive or negative, as long as they appear in the form of a trinomial, the exercise will be called "trinomial".


The first way to factor a trinomial

We will look for two numbers whose product is aβˆ—ca*c and whose sum is bb
We will ask ourselves: which number multiplied by which other number will give us aΓ—ca\times c or ​​c​​c (if aa equals ​​1​​1).
and what plus what would add up to bb.

In fact, we have to find a pair of numbers that meet these two conditions at the same time.

We can plot it as follows:

We can plot it as follows

AC Method:
We will find all the numbers whose products are aΓ—c a\times c and write them down.
Then, we will see which pair of numbers among those we found will result in B B .
The two numbers that meet both conditions are the solutions to the trinomial.

Important

  • If A were different from 1 1 , it would appear before the parentheses and then there would be a multiplication.
  • If any of the solutions or both were negative, we would not add them to the X X but subtract them instead.

Do you know what the answer is?

Let's look at an example of the use of factoring trinomials in the first way.

x2+8x+12x^2+8x+12
Let's find all the numbers whose products are 12 12 (and remember them in negative as well)
we will obtain:
12,112,1
2,62,6
3,43,4
Now let's see which pair of numbers among those we already found will give us a total of 88
The pair that meets both conditions is 2,62,6.
Let's write the factorization:
(x+2)(x+6)(x+2)(x+6)


The second way to factor a trinomial

Let's look at an example of the use of factoring trinomials in the second way:

x2+4x+4=x^2+4x+4=

Let's find our parameters:
aaΒ Β  Β The coefficient of the first term 11
bbΒ  Β The coefficient of the second term 44
cc Β The constant termΒ 44

First, we will place them in the formula with the plus sign and it will give us:
βˆ’4+42βˆ’4Γ—1Γ—42Γ—1=\frac{-4+\sqrt{4^2-4\times 1\times 4}}{2\times 1}=
βˆ’4+16βˆ’162=\frac{-4+\sqrt{16-16}}{2}=
βˆ’4+02=\frac{-4+\sqrt{0}}{2}=
βˆ’42=βˆ’2\frac{-4}{2}=-2
We will place them in the formula with the minus sign and we will get:
βˆ’4βˆ’02=\frac{-4-\sqrt{0}}{2}=
βˆ’42=βˆ’2-\frac{4}{2}=-2

We get the same answer.
The factorization is:
(xβˆ’2)(xβˆ’2)(x-2)(x-2)


Examples and exercises with solutions for factoring trinomials

Exercise #1

x2+6x+9=0 x^2+6x+9=0

What is the value of X?

Video Solution

Step-by-Step Solution

The equation in the problem is:

x2+6x+9=0 x^2+6x+9=0 We want to solve this equation using factoring,

First, we'll check if we can factor out a common factor, but this is not possible, since there is no common factor for all three terms on the left side of the equation, we can identify that we can factor the expression on the left side using the quadratic formula for a trinomial squared, however, we prefer to factor it using the factoring method according to trinomials, let's refer to the search for Factoring by trinomials:

Let's note that the coefficient of the squared term (the term with the second power) is 1, so we can try to perform factoring according to the quick trinomial method: (This factoring is also called "automatic trinomial"),

But before we do this in the problem - let's recall the general rule for factoring by quick trinomial method:

The rule states that for the algebraic quadratic expression of the general form:

x2+bx+c x^2+bx+c We can find a factorization in the form of a product if we can find two numbers m,n m,\hspace{4pt}n such that the following conditions are met (conditions of the quick trinomial method):

{mβ‹…n=cm+n=b \begin{cases} m\cdot n=c\\ m+n=b \end{cases} If we can find two such numbers m,n m,\hspace{4pt}n then we can factor the general expression mentioned above into the form of a product and present it as:

x2+bx+c↓(x+m)(x+n) x^2+bx+c \\ \downarrow\\ (x+m)(x+n) which is its factored form (product factors) of the expression,

Let's return now to the equation in the problem that we received in the last stage after arranging it:

x2+6x+9=0 x^2+6x+9=0 Let's note that the coefficients from the general form we mentioned in the rule above:

x2+bx+c x^2+bx+c are:{c=9b=6 \begin{cases} c=9 \\ b=6 \end{cases} where we didn't forget to consider the coefficient together with its sign,

Let's continue, we want to factor the expression on the left side into factors according to the quick trinomial method, above, so we'll look for a pair of numbers m,n m,\hspace{4pt}n that satisfy:

{mβ‹…n=9m+n=6 \begin{cases} m\cdot n=9\\ m+n=6 \end{cases} We'll try to identify this pair of numbers through logical thinking and using our knowledge of the multiplication table, we'll start from the multiplication between the two required numbers m,n m,\hspace{4pt}n that is - from the first row of the pair of requirements we mentioned in the last stage:

mβ‹…n=9 m\cdot n=9 We identify that their product needs to give a positive result, and therefore we can conclude that their signs are identical,

Next, we'll consider the factors (integers) of the number 9, and from our knowledge of the multiplication table we can know that there are only two possibilities for such factors: 3 and 3, or 9 and 1, as we previously concluded that their signs must be identical, a quick check of the two possibilities regarding the fulfillment of the second condition:

m+n=6 m+n=6 will lead to a quick conclusion that the only possibility for fulfilling both of the above conditions together is:

3,3 3,\hspace{4pt}3 That is - for:

m=3,n=3 m=3,\hspace{4pt}n=3 (It doesn't matter which one we call m and which one we call n)

It is satisfied that:

{3β€Ύβ‹…3β€Ύ=93β€Ύ+3β€Ύ=6 \begin{cases} \underline{3}\cdot \underline{3}=9\\ \underline{3}+\underline{3}=6 \end{cases} From here - we understood what the numbers we are looking for are and therefore we can factor the expression on the left side of the equation in question and present it as a product:

x2+6x+9↓(x+3)(x+3) x^2+6x+9 \\ \downarrow\\ (x+3)(x+3)

In other words, we performed:

x2+bx+c↓(x+m)(x+n) x^2+bx+c \\ \downarrow\\ (x+m)(x+n)

If so, we have factored the quadratic expression on the left side of the equation into factors using the quick trinomial method, and the equation is:

x2+6x+9=0↓(x+3)(x+3)=0(x+3)2=0 x^2+6x+9=0 \\ \downarrow\\ (x+3)(x+3)=0\\ (x+3)^2=0\\ Where in the last stage we noticed that in the expression on the left side the term:

(x+3) (x+3)

multiplies itself and therefore the expression can be written as a squared term:

(x+3)2 (x+3)^2

Now that the expression on the left side has been factored into a product form (in this case not just a product but actually a power form) we will continue to quickly solve the equation we received:

(x+3)2=0 (x+3)^2=0

Let's pay attention to a simple fact, on the left side there is a term raised to the second power, and on the right side the number 0,

and only 0 squared (to the second power) will give the result 0, so we get that the equation equivalent to this equation is the equation:

x+3=0 x+3=0 (In the same way we could have operated algebraically in a pure form and taken the square root of both sides of the equation, we'll discuss this in a note at the end)

We'll solve this equation by moving the free number to the other side and we'll get that the only solution is:

x=βˆ’3 x=-3 Let's summarize then the stages of solving the quadratic equation using the quick trinomial factoring method, we got that:

x2+6x+9=0↓(x+3)(x+3)=0(x+3)2=0↓x+3=0x=βˆ’3 x^2+6x+9=0 \\ \downarrow\\ (x+3)(x+3)=0\\ (x+3)^2=0\\ \downarrow\\ x+3=0\\ x=-3 Therefore, the correct answer is answer C.

Note:

We could have reached the final equation by taking the square root of both sides of the equation, however - taking a square root involves considering two possibilities: positive and negative (it's enough to consider this only on one side, as described in the calculation below), that is, we could have performed:

(x+3)2=0/↓(x+3)2=Β±0x+3=Β±0x+3=0 (x+3)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{8pt}}\\ \downarrow\\ \sqrt{(x+3)^2}=\pm\sqrt{0} \\ x+3=\pm0\\ x+3=0 Where on the left side the root (which is a half power) and the second power canceled each other out (this follows from the law of powers for power over power), and on the right side the root of 0 is 0, and we considered two possibilities positive and negative (this is the plus-minus sign indicated) except that the sign (which is actually multiplication by one or minus one) does not affect 0 which remains 0 in both cases, and therefore we reached the same equation we reached with logical and unambiguous thinking earlier - in the solution above,

In any other case where on the right side was a number different from 0, we could have solved only by taking the root etc. and considering the two positive and negative possibilities which would then give two different possibilities for the solution.

Answer

3-

Exercise #2

Complete the equation:

(x+3)(x+☐)=x2+5x+6 (x+3)(x+\textcolor{red}{☐})=x^2+5x+6

Video Solution

Step-by-Step Solution

Let's simplify the expression given in the left side:

(x+3)(x+?) (x+3)(x+\textcolor{purple}{\boxed{?}}) For ease of calculation we will replace the square with the question mark (indicating the missing part that needs to be completed) with the letter k \textcolor{purple}{k} , meaning we will perform the substitution:

(x+3)(x+?)=x2+5x+6↓(x+3)(x+k)=x2+5x+6 (x+3)(x+\textcolor{purple}{\boxed{?}})=x^2+5x+6 \\ \downarrow\\ (x+3)(x+\textcolor{purple}{k})=x^2+5x+6 \\ Next, we will expand the parentheses using the expanded distribution law:

(a+b)(c+d)=ac+ad+bc+bd (\textcolor{red}{a}+\textcolor{blue}{b})(c+d)=\textcolor{red}{a}c+\textcolor{red}{a}d+\textcolor{blue}{b}c+\textcolor{blue}{b}d Let's note that in the formula template for the distribution law mentioned we assume by default that the operation between the terms inside the parentheses is addition, so we won't forget of course that the sign preceding the term is an integral part of it, and we will also apply the laws of sign multiplication and thus we can represent any expression in parentheses, which we expand using the aforementioned formula, first, as an expression where addition is performed between all terms (if necessary),

Therefore, we will first represent each of the expressions in parentheses in the multiplication on the left side as an expression where addition exists:

(x+3)(x+k)=x2+5x+6↓(x+(+3))(x+(+k))=x2+5x+6 (x+3)(x+\textcolor{purple}{k})=x^2+5x+6 \\ \downarrow\\ \big(x+(+3)\big)\big(x+(\textcolor{purple}{+k})\big)=x^2+5x+6 \\ Now for convenience, let's write down again the expanded distribution law mentioned earlier:

(a+b)(c+d)=ac+ad+bc+bd (\textcolor{red}{a}+\textcolor{blue}{b})(c+d)=\textcolor{red}{a}c+\textcolor{red}{a}d+\textcolor{blue}{b}c+\textcolor{blue}{b}d And we'll apply it to our problem:

(x+(+3))(x+(+k))=x2+5x+6↓xβ‹…x+xβ‹…(+k)+(+3)β‹…x+(+3)β‹…(+k)=x2+5x+6 \big (\textcolor{red}{x}+\textcolor{blue}{(+3)}\big)\big(x+(+\textcolor{purple}{k})\big)=x^2+5x+6 \\ \downarrow\\ \textcolor{red}{x}\cdot x +\textcolor{red}{x}\cdot (+\textcolor{purple}{k})+\textcolor{blue}{(+3)}\cdot x +\textcolor{blue}{(+3)}\cdot (+\textcolor{purple}{k})=x^2+5x+6 \\ We'll continue and apply the laws of multiplication signs, remembering that multiplying expressions with identical signs will yield a positive result, and multiplying expressions with different signs will yield a negative result:

xβ‹…x+xβ‹…(+k)+(+3)β‹…x+(+3)β‹…(+k)=x2+5x+6↓x2+kx+3x+3k=x2+5x+6 \textcolor{red}{x}\cdot x +\textcolor{red}{x}\cdot (+\textcolor{purple}{k})+\textcolor{blue}{(+3)}\cdot x +\textcolor{blue}{(+3)}\cdot (+\textcolor{purple}{k})=x^2+5x+6 \\ \downarrow\\ x^2+\textcolor{purple}{k}x+3x+3\textcolor{purple}{k}=x^2+5x+6 \\ Now, we want to present the expression on the left side in a form identical to the expression on the right side, that is - as a sum of three terms with different exponents: second power (squared), first power, and zero power (i.e., the free number - not dependent on x). To do this - we will factor out the part of the expression on the left side where the terms are in the first power:

x2+kx+3xβ€Ύ+3k=x2+5x+6↓x2+(k+3)xβ€Ύ+3k=x2+5x+6 x^2+\underline{\textcolor{purple}{k}x+3x}+3\textcolor{purple}{k}=x^2+5x+6 \\ \downarrow\\ x^2+\underline{(\textcolor{purple}{k}+3)x}+3\textcolor{purple}{k}=x^2+5x+6

Now in order for equality to hold - we require that the coefficient of the first-power term on both sides of the equation be identical and at the same time - we require that the free term on both sides of the equation be identical as well:

x2+(k+3)β€Ύβ€Ύx+3kβ€Ύβ€Ύβ€Ύ=x2+5β€Ύβ€Ύx+6β€Ύβ€Ύβ€Ύ x^2+\underline{\underline{(\textcolor{purple}{k}+3)}}x+\underline{\underline{\underline{3\textcolor{purple}{k}}}}=x^2+\underline{\underline{5}}x+\underline{\underline{\underline{6}}} In other words, we require that:

{k+3=5β†’k=23k=6β†’k=2 \begin{cases} \textcolor{purple}{k}+3=5\rightarrow\boxed{\textcolor{purple}{k}=2}\\ 3\textcolor{purple}{k}=6\rightarrow\boxed{\textcolor{purple}{k}=2} \end{cases}

Let's summarize the solution steps:

(x+3)(x+?)=x2+5x+6↔?=k↓(x+3)(x+k)=x2+5x+6↓x2+kx+3xβ€Ύ+3k=x2+5x+6↓x2+(k+3)xβ€Ύ+3k=x2+5x+6↓x2+(k+3)β€Ύβ€Ύx+3kβ€Ύβ€Ύβ€Ύ=x2+5β€Ύβ€Ύx+6β€Ύβ€Ύβ€Ύ{k+3=5β†’k=23k=6β†’k=2β†’?=2 (x+3)(x+\textcolor{purple}{\boxed{?}})=x^2+5x+6 \leftrightarrow\textcolor{red}{\boxed{\textcolor{purple}{\boxed{?}}=\textcolor{purple}{k}}} \\ \downarrow\\ (x+3)(x+\textcolor{purple}{k})=x^2+5x+6 \\ \downarrow\\ x^2+\underline{\textcolor{purple}{k}x+3x}+3\textcolor{purple}{k}=x^2+5x+6 \\ \downarrow\\ x^2+\underline{(\textcolor{purple}{k}+3)x}+3\textcolor{purple}{k}=x^2+5x+6\\ \downarrow\\ x^2+\underline{\underline{(\textcolor{purple}{k}+3)}}x+\underline{\underline{\underline{3\textcolor{purple}{k}}}}=x^2+\underline{\underline{5}}x+\underline{\underline{\underline{6}}} \\ \begin{cases} \textcolor{purple}{k}+3=5\rightarrow\boxed{\textcolor{purple}{k}=2}\\ 3\textcolor{purple}{k}=6\rightarrow\boxed{\textcolor{purple}{k}=2} \end{cases} \\ \textcolor{red}{\bm{\rightarrow}\boxed{\textcolor{purple}{\boxed{?}}=\textcolor{purple}{2}}} Therefore, the missing expression is the number 2 2 meaning - the correct answer is a'.

Answer

2

Exercise #3

How many solutions does the equation have?

x4+12x3+36x2=0 x^4+12x^3+36x^2=0

Video Solution

Step-by-Step Solution

Let's solve the given equation:

x4+12x3+36x2=0 x^4+12x^3+36x^2=0 We note that it is possible to factor the expression which is in the left side of the given equation, this is done by taking out the common factor x2 x^2 which is the greatest common factor of the numbers and letters in the expression:

x4+12x3+36x2=0↓x2(x2+12x+36)=0 x^4+12x^3+36x^2=0 \\ \downarrow\\ x^2(x^2+12x+36) =0 We will focus on the left side of the equation and then on the right side (the number 0).

Since the only way to get the result 0 from a product is to multiply by 0, at least one of the expressions in the product on the left side, must be equal to zero,

Meaning:

x2=0/x=0 x^2=0\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\\ \boxed{x=0}

Or:

x2+12x+36=0 x^2+12x+36=0 In order to find the additional solutions to the equation we must solve the equation:

Note that the first coefficient is 1, so we can try to solve it using the trinomial formula.

However, we can factor, in this case, also using the short multiplication formula for a binomial:

(a+b)2=a2+2ab+b2 (\textcolor{red}{a}+\textcolor{blue}{b})^2 = \textcolor{red}{a}^2+2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2 The reason for trying factoring in this approach is that we can identify in the left side of the equation we got in the last step, that the two terms which are in the far sides (meaning the term in the first position - it is the squared term and the term in the zero position - it is the free number in the expression) can be presented (simply) as a squared term:

x2+12x+36=0↓x2+12x+62=0 x^2+12x+36=0 \\ \downarrow\\ x^2+12x+6^2=0

Equating the expression on the left side in the equation:

↓x2+12x+62 \downarrow\\ x^2+12x+6^2

To the expression on the right side in the short formula above:

a2+2ab+b2 \textcolor{red}{a}^2+2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2

The conclusion from this is that what remains to check is whether the middle term in the equation matches the middle term in the short multiplication formula above, meaning - after identifying a a b b which are both in the first position in the short multiplication formula above in which a a and b b we check if the middle term in the expression in the left side of the equation can be presented as 2β‹…aβ‹…b 2\cdot a \cdot b So, we start by presenting the equation of the short formula to the given expression:

a2+2abβ€Ύ+b2⇔x2+12xβ€Ύ+62 \textcolor{red}{a}^2+\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2 \Leftrightarrow \textcolor{red}{x}^2+\underline{12x}+\textcolor{blue}{6}^2 And indeed it holds that:

2β‹…xβ‹…6=12x 2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{6}=12x Meaning the middle term in the expression in the equation indeed matches the form of the middle term in the short multiplication formula (highlighted with a line below), mathematically:

x2+12xβ€Ύ+62=0x2+2β‹…xβ‹…6β€Ύ+62=0↓(x+6)2=0 x^2+\underline{12x}+6^2=0 \\ \textcolor{red}{x}^2+\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{6}}+\textcolor{blue}{6}^2=0\\ \downarrow\\ (\textcolor{red}{x}+\textcolor{blue}{6})^2=0 We can now remember that a real root can be calculated only for a positive number or for the number zero (since it is not possible to get a negative number from squaring a real number itself), and therefore for an equation there are two real solutions (or one solution) only if:

Next we note that if: (x+6)2=0/x+6=0x=βˆ’6 (x+6)^2=0\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\\ x+6=0\\ \boxed{x=-6} then the only solution to the equation is:

x4+12x3+36x2=0↓x2(x2+12x+36)=0↓x2=0β†’x=0x2+12x+36=0x2+2β‹…2β‹…6+62=0β†’(x+6)2=0β†’x=βˆ’6↓x=0,βˆ’6 x^4+12x^3+36x^2=0 \\ \downarrow\\ x^2(x^2+12x+36) =0 \\ \downarrow\\ x^2=0\rightarrow\boxed{x=0} \\ x^2+12x+36=0\\ x^2+2\cdot2\cdot6+6^2=0\\ \rightarrow(x+6)^2=0\rightarrow\boxed{x=-6}\\ \downarrow\\ \boxed{x=0,-6} Therefore, we can summarize what was explained using the following:

In the quadratic equation:

ax2+bx+c=0 ax^2+bx+c =0 in which the coefficients are substituted and the discriminant is calculated:

a,b a,b If it holds:

a.x1,2=βˆ’bΒ±b2βˆ’4ac2a x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a} :

There is no (real) solution to the equation.

b.Ξ” \Delta :

There exists a single (real) solution to the equation.

c.x1,2=βˆ’bΒ±Ξ”2aβŸ·Ξ”=b2βˆ’4ac x_{1,2}=\frac{-b\pm\sqrt{\textcolor{orange}{\Delta}}}{2a} \longleftrightarrow\textcolor{orange}{\Delta}=b^2-4ac :

There exist two (real) solutions to the equation.

Now let's return to the given equation and extract from it the coefficients:

Ξ”β‰₯0 \Delta\geq0 We continue and calculate Ξ”=0 \Delta=0 :

x=βˆ’b2a x=-\frac{b}{2a} Therefore for the quadratic equation that we solved, one (real) solution,

and in combination with the solution ax2+bx+c=0 ax^2+bx+c =0 (the additional solution we found for the given equation which is indicated in the first step after factoring using the common factor),

Therefore we get that for the given equation:

Ξ”=b2βˆ’4ac \Delta=b^2-4ac

two real solutions.

Answer

Two solutions

Exercise #4

Solve the following equation:

4x2βˆ’14xβˆ’8=0 4x^2-14x-8=0

Video Solution

Step-by-Step Solution

Let's solve the given equation:

4x2βˆ’14xβˆ’8=0 4x^2-14x-8=0

Instead of dividing both sides of the equation by the common factor of all terms in the equation (which is the number 2), we will choose to factor it out of the parentheses:

4x2βˆ’14xβˆ’8=02(2x2βˆ’7xβˆ’4)=0 4x^2-14x-8=0 \\ 2(2x^2-7x-4)=0

From here we remember that the result of multiplication between expressions will yield 0 only if at least one of the multiplying expressions equals zero,

However, the first factor in the expression we got is the number 2, which is obviously different from zero, therefore:

2x2βˆ’7xβˆ’4=0 2x^2-7x-4 =0 Now let's note that the coefficient of the quadratic term (squared) is different from 1, we can solve the resulting equation of course using the quadratic formula, but we prefer, for the sake of skill improvement, to continue and factor the expression on the left side, this we will do using general trinomial factoring (actually - grouping method),

Similar to the quick factoring method by trinomial (which is actually a special case of the general trinomial factoring) we will look for a pair of numbers m,n m,\hspace{2pt}n whose product should give the product of the coefficient of the quadratic term and the free term in the general expression:

ax2+bx+c ax^2+bx+c and their sum, the coefficient of the term in the first power, that is in the general expression mentioned, if so - we will look for a pair of numbers m,n m,\hspace{2pt}n that satisfy:

mβ‹…n=aβ‹…cm+n=b m\cdot n=a\cdot c\\ m+n=b Once we find the pair of numbers that satisfy both conditions mentioned (if indeed such can be found) we will separate the coefficient of the term in the first power accordingly and factor by grouping,

Let's return then to the problem and demonstrate:

In the equation:

2x2βˆ’7xβˆ’4=0 2x^2-7x-4 =0 We will look for a pair of numbers m,n m,\hspace{2pt}n that satisfy:

mβ‹…n=2β‹…(βˆ’4)m+n=βˆ’7↓mβ‹…n=βˆ’8m+n=βˆ’7 m\cdot n=2\cdot (-4)\\ m+n=-7 \\ \downarrow\\ m\cdot n=-8\\ m+n=-7 \\ We'll continue, identical to what's done in the quick factoring method by trinomial,

From the first requirement mentioned, that is - from the multiplication, let's note that the product of the numbers we're looking for should yield a negative result and therefore we can conclude that the two numbers have different signs, this is according to multiplication laws, and now we'll remember that the possible factors of the number 8 are 2 and 4 or 8 and 1, fulfilling the second requirement mentioned, along with the fact that the signs of the numbers we're looking for are different from each other will lead to the conclusion that the only possibility for the two numbers we're looking for is:

{m=βˆ’8n=1 \begin{cases} m=-8\\ n=1 \end{cases} From here, unlike the quick factoring method by trinomial (where this step is actually skipped and factored directly, but it definitely exists), we will separate the coefficient of the term in the first power according to the pair of numbers m,n m,\hspace{2pt}n we found:

2x2βˆ’7xβ€Ύβˆ’4=0↓2x2(βˆ’8+1)xβ€Ύβˆ’4=0↓2x2βˆ’8x+xβ€Ύβˆ’4=0 2x^2\underline{\textcolor{blue}{-7x}}-4 =0\\ \downarrow\\ 2x^2\underline{\textcolor{blue}{(-8+1)x}}-4 =0\\ \downarrow\\ 2x^2\underline{\textcolor{blue}{-8x+x}}-4 =0\\ We'll continue - in the next step we will factor by grouping:

We will refer to two groups of terms, so that in each group there is one term in the first power (the choice of groups doesn't matter - as long as this condition is maintained), in each group - we will factor out a common factor so that inside the parentheses, in both groups, we get the same expression:

2x2βˆ’8x+xβˆ’4=0↓2x(xβˆ’4)+1(xβˆ’4)=0 \textcolor{green}{2x^2-8x}\textcolor{red}{+x-4} =0\\ \downarrow\\ \textcolor{green}{2x(x-4)}\textcolor{red}{+1(x-4)} =0\\ (In this case, in the second group - which is marked in red, it was not possible to factor further, so we settled for factoring out the number 1 as a common factor for emphasis),

We'll continue, now let's note that the expression in parentheses in both groups is identical and therefore we can refer to it as a common factor for both groups (which is a binomial) and factor it out of the parentheses:

2x(xβˆ’4)β€Ύ+1(xβˆ’4)β€Ύ=0↓(xβˆ’4)β€Ύ(2x+1)=0 \textcolor{green}{2x}\underline{(x-4)}\textcolor{red}{+1}\underline{(x-4)}=0\\ \downarrow\\ \underline{(x-4)}(\textcolor{green}{2x}\textcolor{red}{+1})=0 We have thus obtained a factored expression on the left side,

Let's summarize this factoring technique:

2x2βˆ’7xβˆ’4=0{mβ‹…n=2β‹…(βˆ’4)=βˆ’8m+n=βˆ’7↔m,n=?↓m=!βˆ’8n=!1↓2x2βˆ’7xβ€Ύβˆ’4=0↓2x2βˆ’8x+xβ€Ύβˆ’4=0↓2x2βˆ’8x+xβˆ’4=0↓2x(xβˆ’4)β€Ύ+1(xβˆ’4)β€Ύ=0↓(xβˆ’4)β€Ύ(2x+1)=0 2x^2-7x-4 =0 \\ \begin{cases} m\cdot n=2\cdot (-4)=-8\\ m+n=-7 \\\end{cases}\leftrightarrow m,\hspace{2pt}n=?\\ \downarrow\\ m\stackrel{!}{= }-8\\ n\stackrel{!}{= }1\\ \downarrow\\ 2x^2\underline{\textcolor{blue}{-7x}}-4 =0\\ \downarrow\\ 2x^2\underline{\textcolor{blue}{-8x+x}}-4 =0\\ \downarrow\\ \textcolor{green}{2x^2-8x}\textcolor{red}{+x-4} =0\\ \downarrow\\ \textcolor{green}{2x}\underline{(x-4)}\textcolor{red}{+1}\underline{(x-4)}=0\\ \downarrow\\ \underline{(x-4)}(\textcolor{green}{2x}\textcolor{red}{+1})=0 This technique is very important - it is recommended to review and sharpen the understanding of it,

Let's continue solving the equation, we got:

(xβˆ’4)(2x+1)=0 (x-4)(2x+1)=0

From here we remember that the result of multiplication between expressions will yield 0 only if at least one of the multiplying expressions equals zero,

Therefore we will get two simple equations and solve them, in the usual way, by isolating the unknown in the side (by transferring sides and division/multiplication of both sides if necessary):

xβˆ’4=0x=4 x-4=0\\ \boxed{x=4} Or :

2x+1=02x=βˆ’1/:2x=βˆ’12 2x+1=0\\ 2x=-1\text{/}:2\\ \boxed{x=-\frac{1}{2}}

Let's summarize then the solution of the equation:

4x2βˆ’14xβˆ’8=02(2x2βˆ’7xβˆ’4)=0↓2x2βˆ’7xβˆ’4=0{mβ‹…n=2β‹…(βˆ’4)=βˆ’8m+n=βˆ’7↔m,n=?↓m=!βˆ’8n=!1↓2x2βˆ’8x+xβˆ’4=0↓2x(xβˆ’4)β€Ύ+1β‹…(xβˆ’4)β€Ύ=0(xβˆ’4)β€Ύ(2x+1)=0↓xβˆ’4=0β†’x=42x+1=0β†’x=βˆ’12↓x=4,βˆ’12 4x^2-14x-8=0 \\ 2(2x^2-7x-4)=0 \\ \downarrow\\ 2x^2-7x-4 =0 \\ \begin{cases} m\cdot n=2\cdot (-4)=-8\\ m+n=-7 \\\end{cases}\leftrightarrow m,\hspace{2pt}n=?\\ \downarrow\\ m\stackrel{!}{= }-8\\ n\stackrel{!}{= }1\\ \downarrow\\ 2x^2\textcolor{blue}{-8x+x}-4 =0 \\ \downarrow\\ 2x\underline{(x-4)}+1\cdot\underline{(x-4)}=0\\ \underline{(x-4)}(2x+1)=0\\ \downarrow\\ x-4=0\rightarrow\boxed{x=4}\\ 2x+1=0\rightarrow\boxed{x=-\frac{1}{2}}\\ \downarrow\\ \boxed{x=4,-\frac{1}{2}} Therefore the correct answer is answer a.

Answer

4,βˆ’12 4,-\frac{1}{2}

Exercise #5

A rectangle has an area of

m2+4mβˆ’12 m^2+4m-12 cmΒ²

and a length of m+6 m+6 cm.

Is the rectangle's perimeter 16 cm?

Video Solution

Step-by-Step Solution

First, let's consider the rectangle ABCD ABCD :

(Sketch - indicating the given information about AB on it)

Let's continue and write down the information about the area of the rectangle and the length of the given side in mathematical form:

{SABCD=m2+4mβˆ’12AB=m+6 \begin{cases} \textcolor{red}{S_{ABCD}}= m^2+4m-12 \\ \textcolor{blue}{AB}=m+6\\ \end{cases} (We'll use colors here for clarity of the solution later)

Now let's remember the fact that the area of a rectangle whose side lengths (adjacent) are:

a,b a,\hspace{2pt}b is:

S=aβ‹…b S_{\boxed{\hspace{6pt}}}=a\cdot b and therefore the area of the rectangle in the problem (according to the sketch we established at the beginning of the solution) is:

SABCD=ABβ‹…AD S_{ABCD}=AB\cdot AD We can now substitute the previously mentioned data into this expression for area to get the equation (for understanding - refer to the marked colors and the data mentioned earlier accordingly):

SABCD=ABβ‹…AD↓m2+4mβˆ’12=(m+6)β‹…AD \textcolor{red}{ S_{ABCD}}=\textcolor{blue}{AB}\cdot AD \\ \downarrow\\ \boxed{ \textcolor{red}{ m^2+4m-12}=\textcolor{blue}{(m+6)}\cdot AD}

Now, let's pause for a moment and ask what our goal is?

Our goal is, of course, to obtain the algebraic expression for the perimeter of the rectangle (in terms of m), for this we remember that the perimeter of a rectangle is the sum of the lengths of all its sides (let's denote by PABCD P_{ABCD} the perimeter of the rectangle),

In addition, we remember that the opposite sides in a rectangle are equal to each other and therefore the perimeter of the given rectangle is:

PABCD=AB+AD+BC+DC{AB=DCAD=BC↓PABCD=2β‹…(AB+AD) P_{ABCD}=AB+AD+BC+DC\\ \begin{cases} AB=DC\\ AD=BC \end{cases}\\ \downarrow\\ \boxed{\textcolor{purple}{P_{ABCD}=2\cdot(AB+AD)} } However, we already have the algebraic expression for the rectangle's side AB AB (from the given information in the problem, previously marked in blue), and therefore all we need in order to calculate the perimeter of the rectangle is the algebraic expression (in terms of m) for the length of side AD AD ,

Let's return then to the equation we reached in the step before considering the perimeter (highlighted with a square around the equation) and isolate AD AD from it, this we'll do by dividing both sides of the equation by the algebraic expression that is the coefficient of AD AD , that is by:(m+6) (m+6) :

(m+6)β‹…AD=m2+4mβˆ’12/:(m+6)↓AD=m2+4mβˆ’12m+6 \boxed{ \textcolor{red}{\textcolor{blue}{(m+6)}\cdot AD= m^2+4m-12}} \hspace{4pt}\text{/:}(m+6)\\ \downarrow\\ AD=\frac{m^2+4m-12}{m+6} Let's continue and simplify the algebraic fraction we got, we'll do this easily by factoring the numerator of the fraction:

m2+4mβˆ’12 m^2+4m-12 We'll use quick trinomial factoring for this (to review the rules of quick trinomial factoring) and we get:

m2+4mβˆ’12↔{?β‹…?=βˆ’12?+?=4 ↓(m+6)(mβˆ’2) m^2+4m-12\leftrightarrow\begin{cases} \boxed{?}\cdot\boxed{?}=-12\\ \boxed{?}+\boxed{?}=4\ \end{cases}\\ \downarrow\\ (m+6)(m-2) and therefore (we'll return to the expression for AD AD ):

AD=m2+4mβˆ’12m+6↓AD=(m+6)(mβˆ’2)m+6↓AD=mβˆ’2 AD=\frac{m^2+4m-12}{m+6} \\ \downarrow\\ AD=\frac{(m+6)(m-2)}{m+6}\\ \downarrow\\ \boxed{AD=m-2} In the final stage, after we factored the numerator of the fraction, we simplified the fraction,

(Sketch - with the found length of AD)

Let's return then to the expression for the perimeter of the rectangle, which we got earlier and substitute into it the algebraic expressions for the lengths of the rectangle's sides that we got, then we'll simplify the resulting expression:

PABCD=2(AB+AD)AB=m+6AD=mβˆ’2↓PABCD=2(m+6+mβˆ’2)PABCD=2(2m+4)PABCD=4m+8 \boxed{\textcolor{purple}{P_{ABCD}=2(AB+AD)} } \\ AB=m+6\\ AD=m-2\\ \downarrow\\ P_{ABCD}=2(m+6+m-2) \\ P_{ABCD}=2(2m+4) \\ \boxed{P_{ABCD}=4m+8} (length units)

We have thus found the expression for the perimeter of the rectangle in terms of m,

Now let's try to answer the question asked:

Is it possible that the perimeter of the rectangle is 16 length units?

In other words, mathematically- does there exist an m for which:

PABCD=16 P_{ABCD}=16 ?

To answer this question, we'll need to perform two steps:

a. Find the value of m for which the stated condition is met.

b. Check if the value of m that meets the requirement is logical in terms of all the problem data (which may be limited in terms of m).

Let's start with a:

Let's find the value of m that meets the stated requirement,

For this, we'll recall the expression for the perimeter that we found in the previous step (enclosed in a frame), and the stated requirement, then we'll demand that the requirement is met and solve the resulting equation:

{PABCD=4m+8PABCD=16↓4m+8=164m=8/:4m=2 \begin{cases} \boxed{P_{ABCD}=4m+8}\\ P_{ABCD}=16 \end{cases}\\ \downarrow\\ 4m+8=16\\ 4m=8\hspace{6pt}\text{/:}4\\ \boxed{m=2} We have thus found the value of m that meets the requirement regarding the perimeter of the rectangle.

b. Now let's check if for the value of m we found (which meets the requirement regarding the perimeter) we get logical values,

We'll consider that the realistic values in the problem that depend on m are the area and the lengths of the sides, and therefore can only receive positive values (i.e., greater than 0),

For this, we'll recall the various expressions for the area and the lengths of the sides and we'll also state the question asked here mathematically:

m=?2↔{SABCD=(m+6)(mβˆ’2)AB=m+6AD=mβˆ’2 m\stackrel{?}{= }2\leftrightarrow \begin{cases} \textcolor{red}{S_{ABCD}}=(m+6)(m-2) \\ \textcolor{blue}{AB}=m+6\\ AD=m-2 \end{cases} Where we wrote the expression for the area in its expanded form,

We'll use substitution and note that for: m=2 m=2 , the length of side AD AD becomes zero, and additionally, as expected - the area of the rectangle also becomes zero, that is mathematically:

m=2↔{SABCD=(2β€Ύ+6)(2β€Ύβˆ’2)=0AB=2β€Ύ+6=8AD=2β€Ύβˆ’2=0 m=2\leftrightarrow \begin{cases} \textcolor{red}{S_{ABCD}}=(\underline{2}+6)(\underline{2}-2)=0 \\ \textcolor{blue}{AB}=\underline{2}+6=8\\ AD=\underline{2}-2=0 \end{cases} Therefore, we can conclude that the solution m=2, which was obtained from the requirement that the perimeter of the rectangle be equal to: 16 length units,

is not possible in the problem, since for this value of m the length of the rectangle's side (and therefore also its area) becomes zero,

But this value is the only value for which the requirement regarding the perimeter of the rectangle is met,

That is - there does not exist an m for which the perimeter of the rectangle is 16 length units,

Therefore, it is not possible that the perimeter of the given rectangle is: 16 length units.

Therefore, the correct answer is answer b.

Answer

No

Check your understanding
Start practice