Solving Trinomials Practice Problems - Step by Step Solutions

Master trinomial factoring with interactive practice problems. Learn to solve quadratic equations by factoring trinomials of the form ax² + bx + c step by step.

📚Practice Solving Trinomials with Expert Guidance
  • Factor trinomials of the form x² + bx + c using the AC method
  • Solve quadratic equations by factoring trinomial expressions
  • Apply trinomial factoring to simplify complex algebraic fractions
  • Master negative coefficient trinomials like -x² - 4x + 8
  • Use the quadratic formula method for difficult trinomial problems
  • Identify and factor trinomials with parameters like ax² + bx + c

Understanding Solving Trinomials

Complete explanation with examples

What is a Trinomial?

Trinomial is a Greek word composed of two components:

  • Tri - is three in Greek
  • Nom - is partition - with unknown to the square (that is, to the second power)

Therefore: The squared trinomial is an algebraic expression composed of three terms, one with the unknown to the square (to the power 22), the second with the unknown without exponentiation, and the third, numbers without unknowns or letters that are different from those carrying the unknown.

For example, the form:

X2+6X+8X^2+6X+8

Is a trinomial. Likewise, the form:

−X2−4X+8-X^2-4X+8

is also, even though some of the coefficients of the unknowns are negative.

And, in a general form:

aX2+bX+caX^2+bX+c

It is a squared trinomial when the letters a,b,ca,b,c : are correct for any number we place in their place, except for 00

  • Note that, placing a 00 in place of the a will nullify the structure of the quadratic form
  • on the other hand, placing a 00 in place of the parameters bb or cc would turn the trinomial expression (three monomials) into one with only 22 monomials.

Quadratic expression breakdown showing that two numbers must multiply to constant term c and add to coefficient b in the standard form x² + bx + c

Detailed explanation

Practice Solving Trinomials

Test your knowledge with 3 quizzes

\( 3x^3-10x^2+7x=0 \)

Examples with solutions for Solving Trinomials

Step-by-step solutions included
Exercise #1

4x2=12x−9 4x^2=12x-9

Step-by-Step Solution

To solve the quadratic equation 4x2=12x−9 4x^2 = 12x - 9 , we begin by rewriting it in standard quadratic form:

4x2−12x+9=04x^2 - 12x + 9 = 0

Here, we compare to the general form ax2+bx+c=0 ax^2 + bx + c = 0 and identify:

  • a=4a = 4
  • b=−12b = -12
  • c=9c = 9

We will now use the quadratic formula:

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substitute in the values for aa, bb, and cc:

x=−(−12)±(−12)2−4⋅4⋅92⋅4x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4 \cdot 4 \cdot 9}}{2 \cdot 4}

Simplify:

x=12±144−1448x = \frac{12 \pm \sqrt{144 - 144}}{8}

x=12±08x = \frac{12 \pm \sqrt{0}}{8}

x=12±08x = \frac{12 \pm 0}{8}

This simplifies further to:

x=128=32x = \frac{12}{8} = \frac{3}{2}

Therefore, the solution to the equation 4x2=12x−9 4x^2 = 12x - 9 is x=32 x = \frac{3}{2} .

Answer:

x=32 x=\frac{3}{2}

Video Solution
Exercise #2

Solve the following problem:

x2+10x=−25 x^2+10x=-25

Step-by-Step Solution

Proceed to solve the given equation:

x2+10x=−25 x^2+10x=-25

First, let's arrange the equation by moving terms:

x2+10x=−25x2+10x+25=0 x^2+10x=-25 \\ x^2+10x+25=0 \\ Note that the expression on the left side can be factored using the perfect square trinomial formula for a binomial squared:

(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

As shown below:

25=52 25=5^2

Therefore, we'll represent the rightmost term as a squared term:

x2+10x+25=0↓x2+10x+52=0 x^2+10x+25=0 \\ \downarrow\\ \textcolor{red}{x}^2+10x+\textcolor{blue}{5}^2=0

Now let's examine again the perfect square trinomial formula mentioned earlier:

(a+b)2=a2+2ab‾+b2 (\textcolor{red}{a}+\textcolor{blue}{b})^2=\textcolor{red}{a}^2+\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2

And the expression on the left side in the equation that we obtained in the last step:

x2+10x‾+52=0 \textcolor{red}{x}^2+\underline{10x}+\textcolor{blue}{5}^2=0

Notice that the terms x2,52 \textcolor{red}{x}^2,\hspace{6pt}\textcolor{blue}{5}^2 indeed match the form of the first and third terms in the perfect square trinomial formula (which are highlighted in red and blue),

However, in order to factor this expression (on the left side of the equation) using the perfect square trinomial formula mentioned, the remaining term must also match the formula, meaning the middle term in the expression (underlined with a line):

(a+b)2=a2+2ab‾+b2 (\textcolor{red}{a}+\textcolor{blue}{b})^2=\textcolor{red}{a}^2+\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2

In other words - we will query whether we can represent the expression on the left side as:

x2+10x‾+52=0↕?x2+2⋅x⋅5‾+52=0 \textcolor{red}{x}^2+\underline{10x}+\textcolor{blue}{5}^2=0 \\ \updownarrow\text{?}\\ \textcolor{red}{x}^2+\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{5}}+\textcolor{blue}{5}^2=0

And indeed it is true that:

2⋅x⋅5=10x 2\cdot x\cdot5=10x

Therefore we can represent the expression on the left side of the equation as a perfect square binomial:

x2+2⋅x⋅5+52=0↓(x+5)2=0 \textcolor{red}{x}^2+2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{5}+\textcolor{blue}{5}^2=0 \\ \downarrow\\ (\textcolor{red}{x}+\textcolor{blue}{5})^2=0

From here we can take the square root of both sides of the equation (and don't forget there are two possibilities - positive and negative when taking the square root of an even power), then we'll easily solve by isolating the variable:

(x+5)2=0/x+5=±0x+5=0x=−5 (x+5)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ x+5=\pm0\\ x+5=0\\ \boxed{x=-5}

Let's summarize the solution of the equation:

x2+10x=−25x2+10x+25=0↓x2+2⋅x⋅5+52=0↓(x+5)2=0↓x+5=0↓x=−5 x^2+10x=-25 \\ x^2+10x+25=0 \\ \downarrow\\ \textcolor{red}{x}^2+2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{5}+\textcolor{blue}{5}^2=0 \\ \downarrow\\ (\textcolor{red}{x}+\textcolor{blue}{5})^2=0 \\ \downarrow\\ x+5=0\\ \downarrow\\ \boxed{x=-5}

Therefore the correct answer is answer C.

Answer:

x=−5 x=-5

Video Solution
Exercise #3

What is the value of x?

x4−x3=2x2 x^4-x^3=2x^2

Step-by-Step Solution

To solve the problem x4−x3=2x2 x^4 - x^3 = 2x^2 , let's proceed as follows:

  • Step 1: Set the equation to zero.
    x4−x3−2x2=0 x^4 - x^3 - 2x^2 = 0
  • Step 2: Factor out the greatest common factor.
    The common factor among all terms is x2 x^2 .
    Factoring out x2 x^2 gives:
    x2(x2−x−2)=0 x^2(x^2 - x - 2) = 0
  • Step 3: Solve the factors.
    This equation breaks into two factors that can be solved separately:
    • x2=0 x^2 = 0
    • x2−x−2=0 x^2 - x - 2 = 0
  • Step 4: Solve x2=0 x^2 = 0 .
    Since x2=0 x^2 = 0 , we get:
    x=0 x = 0
  • Step 5: Solve x2−x−2=0 x^2 - x - 2 = 0 .
    This can be factored further. We look for two numbers that multiply to −2-2 and add up to −1-1.
    These numbers are −2-2 and 11, so we factor as:
    (x−2)(x+1)=0 (x - 2)(x + 1) = 0
  • Step 6: Solve the quadratic factors.
    Set each factor equal to zero:
    • x−2=0⇒x=2 x - 2 = 0 \Rightarrow x = 2
    • x+1=0⇒x=−1 x + 1 = 0 \Rightarrow x = -1

The solutions to the equation x4−x3=2x2 x^4 - x^3 = 2x^2 are x=−1,0,2 x = -1, 0, 2 .

Therefore, the correct answer is:

x=−1,2,0 x = -1, 2, 0

Answer:

x=−1,2,0 x=-1,2,0

Video Solution
Exercise #4

Solve the following problem:

x2=6x−9 x^2=6x-9

Step-by-Step Solution

Proceed to solve the given equation:

x2=6x−9 x^2=6x-9

First, let's arrange the equation by moving terms:

x2=6x−9x2−6x+9=0 x^2=6x-9 \\ x^2-6x+9=0

Note that we can factor the expression on the left side by using the perfect square trinomial formula for a binomial squared:

(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

As shown below:

9=32 9=3^2 Therefore, we'll represent the rightmost term as a squared term:

x2−6x+9=0↓x2−6x+32=0 x^2-6x+9=0 \\ \downarrow\\ \textcolor{red}{x}^2-6x+\textcolor{blue}{3}^2=0

Now let's examine again the perfect square trinomial formula mentioned earlier:

(a−b)2=a2−2ab‾+b2 (\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2

And the expression on the left side in the equation that we obtained in the last step:

x2−6x‾+32=0 \textcolor{red}{x}^2-\underline{6x}+\textcolor{blue}{3}^2=0

Note that the terms x2,32 \textcolor{red}{x}^2,\hspace{6pt}\textcolor{blue}{3}^2 indeed match the form of the first and third terms in the perfect square trinomial formula (which are highlighted in red and blue),

However, in order to factor the expression in question (which is on the left side of the equation) using the perfect square trinomial formula mentioned, the remaining term must also match the formula, meaning the middle term in the expression (underlined with a single line):

(a−b)2=a2−2ab‾+b2 (\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2

In other words - we will query whether we can represent the expression on the left side of the equation as:

x2−6x‾+32=0↕?x2−2⋅x⋅3‾+32=0 \textcolor{red}{x}^2-\underline{6x}+\textcolor{blue}{3}^2=0\\ \updownarrow\text{?}\\ \textcolor{red}{x}^2-\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{3}}+\textcolor{blue}{3}^2=0

And indeed it is true that:

2⋅x⋅3=6x 2\cdot x\cdot3=6x

Therefore we can represent the expression on the left side of the equation as a perfect square binomial:

x2−2⋅x⋅3‾+32=0↓(x−3)2=0 \textcolor{red}{x}^2-\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{3}}+\textcolor{blue}{3}^2=0 \\ \downarrow\\ (\textcolor{red}{x}-\textcolor{blue}{3})^2=0

From here we can take the square root of both sides of the equation (and don't forget that there are two possibilities - positive and negative when taking an even root of both sides of an equation), then we'll easily solve by isolating the variable:

(x−3)2=0/x−3=±0x−3=0x=3 (x-3)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ x-3=\pm0\\ x-3=0\\ \boxed{x=3}

Let's summarize the solution of the equation:

x2=6x−9x2−6x+9=0↓x2−2⋅x⋅3+32=0↓(x−3)2=0↓x−3=0↓x=3 x^2=6x-9 \\ x^2-6x+9=0 \\ \downarrow\\ \textcolor{red}{x}^2-2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{3}+\textcolor{blue}{3}^2=0 \\ \downarrow\\ (\textcolor{red}{x}-\textcolor{blue}{3})^2=0 \\ \downarrow\\ x-3=0\\ \downarrow\\ \boxed{x=3}

Therefore the correct answer is answer C.

Answer:

x=3 x=3

Video Solution
Exercise #5

Solve the following problem:

x2+144=24x x^2+144=24x

Step-by-Step Solution

Proceed to solve the given equation:

x2+144=24x x^2+144=24x

Arrange the equation by moving terms:

x2+144=24xx2−24x+144=0 x^2+144=24x \\ x^2-24x+144=0

Note that we are able to factor the expression on the left side by using the perfect square trinomial formula:

(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

As demonstrated below:

144=122 144=12^2

Therefore, we'll represent the rightmost term as a squared term:

x2−24x+144=0↓x2−24x+122=0 x^2-24x+144=0 \\ \downarrow\\ \textcolor{red}{x}^2-24x+\textcolor{blue}{12}^2=0

Now let's examine once again the perfect square trinomial formula mentioned earlier:

(a−b)2=a2−2ab‾+b2 (\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2

And the expression on the left side in the equation that we obtained in the last step:

x2−24x‾+122=0 \textcolor{red}{x}^2-\underline{24x}+\textcolor{blue}{12}^2=0

Note that the terms x2,122 \textcolor{red}{x}^2,\hspace{6pt}\textcolor{blue}{12}^2 indeed match the form of the first and third terms in the perfect square trinomial formula (which are highlighted in red and blue),

However, in order to factor this expression (which is on the left side of the equation) using the perfect square trinomial formula mentioned, the remaining term must also match the formula, meaning the middle term in the expression (underlined):

(a−b)2=a2−2ab‾+b2 (\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2

In other words - we'll query whether we can represent the expression on the left side of the equation as:

x2−24x‾+122=0↕?x2−2⋅x⋅12‾+122=0 \textcolor{red}{x}^2-\underline{24x}+\textcolor{blue}{12}^2=0\\ \updownarrow\text{?}\\ \textcolor{red}{x}^2-\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{12}}+\textcolor{blue}{12}^2=0

And indeed it is true that:

2⋅x⋅12=24x 2\cdot x\cdot12=24x

Therefore we can represent the expression on the left side of the equation as a perfect square trinomial:

x2−2⋅x⋅12‾+122=0↓(x−12)2=0 \textcolor{red}{x}^2-\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{12}}+\textcolor{blue}{12}^2=0 \\ \downarrow\\ (\textcolor{red}{x}-\textcolor{blue}{12})^2=0

From here we can take the square root of both sides of the equation (and don't forget that there are two possibilities - positive and negative when taking an even root of both sides of an equation), then we'll easily solve by isolating the variable:

(x−12)2=0/x−12=±0x−12=0x=12 (x-12)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ x-12=\pm0\\ x-12=0\\ \boxed{x=12}

Let's summarize the solution of the equation:

x2+144=24xx2−24x+144=0↓x2−2⋅x⋅12+122=0↓(x−12)2=0↓x−12=0↓x=12 x^2+144=24x \\ x^2-24x+144=0 \\ \downarrow\\ \textcolor{red}{x}^2-2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{12}+\textcolor{blue}{12}^2=0 \\ \downarrow\\ (\textcolor{red}{x}-\textcolor{blue}{12})^2=0 \\ \downarrow\\ x-12=0\\ \downarrow\\ \boxed{x=12}

Therefore the correct answer is answer C.

Answer:

x=12 x=12

Video Solution

Frequently Asked Questions

Everything you need to know about Solving Trinomials

What is a trinomial in algebra and how do I identify one?

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A trinomial is an algebraic expression with exactly three terms, typically in the form ax² + bx + c where 'a' is the coefficient of x², 'b' is the coefficient of x, and 'c' is the constant term. Examples include x² + 6x + 8 and -x² - 4x + 8.

How do I factor a trinomial step by step?

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To factor a trinomial like x² + 5x + 6: 1) Find two numbers that multiply to give the constant term (6) and add to give the middle coefficient (5). 2) These numbers are 2 and 3 since 2×3=6 and 2+3=5. 3) Write as (x+2)(x+3).

What's the difference between factoring x² + bx + c and ax² + bx + c?

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When the leading coefficient is 1 (x² + bx + c), find two numbers that multiply to c and add to b. When a ≠ 1 (ax² + bx + c), find two numbers that multiply to ac and add to b, then use grouping or the AC method.

How do I solve quadratic equations using trinomial factoring?

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First factor the trinomial completely, then set each factor equal to zero using the zero product property. For example: x² + 5x + 6 = 0 becomes (x+2)(x+3) = 0, so x = -2 or x = -3.

What do I do when the trinomial has negative coefficients?

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Apply the same factoring rules but pay attention to signs. For x² - 5x + 6, find two negative numbers that multiply to +6 and add to -5: (-2) and (-3). The factored form is (x-2)(x-3).

When should I use the quadratic formula instead of factoring?

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Use the quadratic formula when the trinomial doesn't factor nicely with integer solutions, when the discriminant (b² - 4ac) isn't a perfect square, or when dealing with complex coefficients that make factoring difficult.

How are trinomials used in simplifying algebraic fractions?

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Factor both the numerator and denominator trinomials, then cancel common factors. For example: (x² + 2x + 1)/(x² - x - 2) = (x+1)(x+1)/(x+1)(x-2) = (x+1)/(x-2).

What are the most common mistakes when factoring trinomials?

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Common errors include: mixing up the signs when dealing with negative terms, forgetting to check if the trinomial factors completely, not verifying the factored form by expanding, and incorrectly applying the AC method for trinomials where a ≠ 1.

Continue Your Math Journey

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Suggested Topics to Practice in Advance

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Topics Learned in Later Sections

Methods for Solving a Quadratic FunctionSolution by extracting a rootCompleting the square in a quadratic equation

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