Solving Trinomials - Examples, Exercises and Solutions

Question Types:
Solving Trinomials: Third powerSolving Trinomials: Factoring Out the Greatest Common Factor (GCF)Solving Trinomials: Solving the problem

More than once we have heard the teacher ask in class: "Who knows how to solve a quadratic equation without the formula?" We looked around to see who knew the answer, "Do you know what a trinomial is?" The teacher continued asking. We doubted and thought about what word this term could derive from and what a trinomial is. What does it really do? How does understanding about the trinomial benefit our mathematical knowledge? Does it expand the possibility of having greater mathematical efficiency? Or, in fact, might it be superfluous to include it in the ninth-grade curriculum?

In this article, we will try to answer these questions and even have fun with the properties of the trinomial that will help us quickly solve quadratic equations, to simplify fractions, to multiply and divide, to deal with fractions, even with the common denominator in fractions with variables in the numerator and in the denominator.

Detailed explanation

Suggested Topics to Practice in Advance

  1. The quadratic equation

Practice Solving Trinomials

Question 1

\( 60-16y+y^2=-4 \)

Question 2

\( x^2+10x=-25 \)

Question 3

\( 4x^2=12x-9 \)

Question 4

\( x^2-10x=-16 \)

Question 5

\( x^2+144=24x \)

Examples with solutions for Solving Trinomials

Exercise #1

60−16y+y2=−4 60-16y+y^2=-4

Video Solution

Step-by-Step Solution

Let's solve the given equation:

60−16y+y2=−4 60-16y+y^2=-4 First, let's arrange the equation by moving terms:

60−16y+y2=−460−16y+y2+4=0y2−16y+64=0 60-16y+y^2=-4 \\ 60-16y+y^2+4=0 \\ y^2-16y+64=0 Now, let's note that we can break down the expression on the left side using the short quadratic factoring 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 This is done using the fact that:

64=82 64=8^2 So let's present the outer term on the right as a square:

y2−16y+64=0↓y2−16y+82=0 y^2-16y+64=0 \\ \downarrow\\ \textcolor{red}{y}^2-16y+\textcolor{blue}{8}^2=0 Now let's examine again the short factoring formula we 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 of the equation we got in the last step:

y2−16y‾+82=0 \textcolor{red}{y}^2-\underline{16y}+\textcolor{blue}{8}^2=0 Let's note that the terms y2,82 \textcolor{red}{y}^2,\hspace{6pt}\textcolor{blue}{8}^2 indeed match the form of the first and third terms in the short multiplication formula (which are highlighted in red and blue),

But in order for us to break down the relevant expression (which is on the left side of the equation) using the short formula we mentioned, the match to the short formula must also apply to the remaining term, 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 ask if it's possible to present the expression on the left side of the equation as:

y2−16y‾+82=0↕?y2−2⋅y⋅8‾+82=0 \textcolor{red}{y}^2-\underline{16y}+\textcolor{blue}{8}^2 =0 \\ \updownarrow\text{?}\\ \textcolor{red}{y}^2-\underline{2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{8}}+\textcolor{blue}{8}^2 =0 And indeed it holds that:

2⋅y⋅8=16y 2\cdot y\cdot8=16y So we can present the expression on the left side of the given equation as a difference of two squares:

y2−2⋅y⋅8+82=0↓(y−8)2=0 \textcolor{red}{y}^2-2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{8}+\textcolor{blue}{8}^2=0 \\ \downarrow\\ (\textcolor{red}{y}-\textcolor{blue}{8})^2=0 From here we can take out square roots for the two sides of the equation (remember that there are two possibilities - positive and negative when taking out square roots), we'll solve it easily by isolating the variable on one side:

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

Let's summarize then the solution of the equation:

60−16y+y2=−4y2−16y+64=0↓y2−2⋅y⋅8+82=0↓(y−8)2=0↓y−8=0↓y=8 60-16y+y^2=-4 \\ y^2-16y+64=0 \\ \downarrow\\ \textcolor{red}{y}^2-2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{8}+\textcolor{blue}{8}^2=0 \\ \downarrow\\ (\textcolor{red}{y}-\textcolor{blue}{8})^2=0 \\ \downarrow\\ y-8=0\\ \downarrow\\ \boxed{y=8}

So the correct answer is answer a.

Answer

y=8 y=8

Exercise #2

x2+10x=−25 x^2+10x=-25

Video Solution

Answer

x=−5 x=-5

Exercise #3

4x2=12x−9 4x^2=12x-9

Video Solution

Answer

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

Exercise #4

x2−10x=−16 x^2-10x=-16

Video Solution

Answer

x=2,8 x=2,8

Exercise #5

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

Video Solution

Answer

x=12 x=12

Question 1

\( x^2=6x-9 \)

Question 2

What is the value of x?

\( x^4-x^3=2x^2 \)

Question 3

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

Question 4

Solve for y:

\( y^2+4y+2=-2 \)

Question 5

Solve for x:

\( x^2+32x=-256 \)

Exercise #6

x2=6x−9 x^2=6x-9

Video Solution

Answer

x=3 x=3

Exercise #7

What is the value of x?

x4−x3=2x2 x^4-x^3=2x^2

Video Solution

Answer

x=−1,2,0 x=-1,2,0

Exercise #8

x3=x2+2x x^3=x^2+2x

Video Solution

Answer

x=0,−1,2 x=0,-1,2

Exercise #9

Solve for y:

y2+4y+2=−2 y^2+4y+2=-2

Video Solution

Answer

y=−2 y=-2

Exercise #10

Solve for x:

x2+32x=−256 x^2+32x=-256

Video Solution

Answer

x=−16 x=-16

Question 1

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

Question 2

\( x^3-7x^2+6x=0 \)

Question 3

\( x^3+x^2-12x=0 \)

Exercise #11

3x3−10x2+7x=0 3x^3-10x^2+7x=0

Video Solution

Answer

x=0,1,73 x=0,1,\frac{7}{3}

Exercise #12

x3−7x2+6x=0 x^3-7x^2+6x=0

Video Solution

Answer

x=0,1,6 x=0,1,6

Exercise #13

x3+x2−12x=0 x^3+x^2-12x=0

Video Solution

Answer

x=0,3,−4 x=0,3,-4

Topics learned in later sections

  1. Methods for Solving a Quadratic Function
  2. Completing the square in a quadratic equation