# Factoring Trinomials - Examples, Exercises and Solutions

## I present to you the following trinomial

#### $ax^2+bx+c$

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:

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

## Practice Factoring Trinomials

### Exercise #1

How many solutions does the equation have?

$x^4+12x^3+36x^2=0$

### Step-by-Step Solution

Let's solve the given equation:

$x^4+12x^3+36x^2=0$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 $x^2$ which is the greatest common factor of the numbers and letters in the expression:

$x^4+12x^3+36x^2=0 \\ \downarrow\\ x^2(x^2+12x+36) =0$We continue and refer to the expression on the left side of the equation that we got in the last step which contains algebraic expressions and on the right side the number 0, therefore, 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:

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

Or:

$x^2+12x+36=0$We continue and in order to find the additional solutions to the equation we solve the equation that we got in the last step:

Note that the leading coefficient is 1, so we can (try) to solve it using the quick trinomial formula,

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

$(\textcolor{red}{a}+\textcolor{blue}{b})^2 = \textcolor{red}{a}^2+2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2$The motivation for trying factoring in this approach is the fact that we can identify in the left side expression in the equation we got in the last step, that the two terms which are in the extreme positions (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:

$x^2+12x+36=0 \\ \downarrow\\ x^2+12x+6^2=0$

Equating the expression on the left side in the equation:

$\downarrow\\ x^2+12x+6^2$

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

$\textcolor{red}{a}^2+2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2$

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

$\textcolor{red}{a}^2+\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2 \Leftrightarrow \textcolor{red}{x}^2+\underline{12x}+\textcolor{blue}{6}^2$And indeed it holds that:

$2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{6}=12x$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:

$x^2+\underline{12x}+6^2=0 \\ \textcolor{red}{x}^2+\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{6}}+\textcolor{blue}{6}^2=0\\ \downarrow\\ (\textcolor{red}{x}+\textcolor{blue}{6})^2=0$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: $(x+6)^2=0\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\\ x+6=0\\ \boxed{x=-6}$then the only solution to the equation is:

$x^4+12x^3+36x^2=0 \\ \downarrow\\ x^2(x^2+12x+36) =0 \\ \downarrow\\ x^2=0\rightarrow\boxed{x=0} \\ x^2+12x+36=0\\ x^2+2\cdot2\cdot6+6^2=0\\ \rightarrow(x+6)^2=0\rightarrow\boxed{x=-6}\\ \downarrow\\ \boxed{x=0,-6}$Therefore, we can summarize what was explained in the rule:

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

$a,b$If it holds:

a.$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$:

There is no (real) solution to the equation.

b.$\Delta$:

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

c.$x_{1,2}=\frac{-b\pm\sqrt{\textcolor{orange}{\Delta}}}{2a} \longleftrightarrow\textcolor{orange}{\Delta}=b^2-4ac$:

There exist two (real) solutions to the equation.

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

$\Delta\geq0$We continue and calculate $\Delta=0$:

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

and in combination with the solution $ax^2+bx+c =0$(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:

$\Delta=b^2-4ac$

two real solutions.

Two solutions

### Exercise #2

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

What is the value of X?

3-

### Exercise #3

$x^2-3x+2=0$

What is the value of X?

### Video Solution

$1,2$

### Exercise #4

$x^2-12x+36=0$

What is the value of X?

6

### Exercise #5

$-x^2+13x-14=0$

What is the value of X?

### Video Solution

$x_1=14,x_2=-1$

### Exercise #1

$2x^2+4x-6=0$

What is the value of X?

### Video Solution

$x_1=1,x_2=-3$

### Exercise #2

$x^2-1=0$

### Video Solution

$x=\pm1$

### Exercise #3

$x^2-5x-50=0$

### Video Solution

$x=10,x=-5$

### Exercise #4

$x^2-8x+16=0$

### Video Solution

$x=4$

### Exercise #5

$x^2+x-2=0$

### Video Solution

$(x-1)(x+2)=0$

### Exercise #1

$x^2+9x+20=0$

### Video Solution

$x=-4,x=-5$

### Exercise #2

$x^2-2x-3=0$

### Video Solution

$x=3,x=-1$

### Exercise #3

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

### Video Solution

$x=-3$

### Exercise #4

$x^2-7x+12=0$

### Video Solution

$x=3,x=4$

### Exercise #5

$x^2-19x+60=0$

### Video Solution

$x=15,x=4$