Solving Trinomials - Examples, Exercises and Solutions

Question Types:
Solving Trinomials: Factoring Out the Greatest Common Factor (GCF)Solving Trinomials: Solving the problemSolving Trinomials: Third power

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

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

Question 2

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

Question 3

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

Question 4

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

Question 5

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

Examples with solutions for Solving Trinomials

Exercise #1

x2+10x=−25 x^2+10x=-25

Video Solution

Step-by-Step Solution

Let's 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 \\ Now we notice 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

We can do this using the fact that:

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 we got 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'll ask if 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

Exercise #2

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 #3

x2−10x=−16 x^2-10x=-16

Video Solution

Step-by-Step Solution

Let's solve the given equation:

x2−10x=−16 x^2-10x=-16

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

x2−10x=−16x2−10x+16=0 x^2-10x=-16 \\ x^2-10x+16 =0

Now we notice that the coefficient of the squared term is 1, therefore, we can (try to) factor the expression on the left side using quick trinomial factoring:

Let's look for a pair of numbers whose product equals the free term in the expression, and whose sum equals the coefficient of the first-degree term, meaning two numbers m,n m,\hspace{2pt}n that satisfy:

m⋅n=16m+n=−10 m\cdot n=16\\ m+n=-10\\ From the first requirement, namely - the multiplication, we notice that the product of the numbers we're looking for must yield a positive result, therefore we can conclude that both numbers must have the same sign, according to multiplication rules, and now we'll remember that the possible factors of 16 are the number pairs 4 and 4, 2 and 8, or 16 and 1. Meeting the second requirement, along with the fact that the signs of the numbers we're looking for are identical will lead to the conclusion that the only possibility for the two numbers we're looking for is:

{m=−8n=−2 \begin{cases} m=-8\\ n=-2 \end{cases}

Therefore we'll factor the expression on the left side of the equation to:

x2−10x+16=0↓(x−8)(x−2)=0 x^2-10x+16 =0 \\ \downarrow\\ (x-8)(x-2)=0

From here we'll remember that the product of expressions will yield 0 only if at least one of the multiplied expressions equals zero,

Therefore we'll get two simple equations and solve them by isolating the unknown in each:

x−8=0x=8 x-8=0\\ \boxed{x=8}

or:

x−2=0x=2 x-2=0\\ \boxed{x=2}

Let's summarize the solution of the equation:

x2−10x=−16x2−10x+16=0↓(x−8)(x−2)=0↓x−8=0→x=8x−2=0→x=2↓x=8,2 x^2-10x=-16 \\ x^2-10x+16 =0 \\ \downarrow\\ (x-8)(x-2)=0 \\ \downarrow\\ x-8=0\rightarrow\boxed{x=8}\\ x-2=0\rightarrow\boxed{x=2}\\ \downarrow\\ \boxed{x=8,2}

Therefore the correct answer is answer B.

Answer

x=2,8 x=2,8

Exercise #4

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

Video Solution

Step-by-Step Solution

Let's solve the given equation:

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

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

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

Now let's notice that we can factor the expression on the left side 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

We can do this using the fact that:

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 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 we got in the last step:

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

Notice 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 ask if 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

Exercise #5

x2=6x−9 x^2=6x-9

Video Solution

Step-by-Step Solution

Let's 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

Now let's notice that we can factor the expression on the left side 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

We'll do this using the fact that:

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 we got in the last step:

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

Notice 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'll ask if 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

Question 1

Solve for y:

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

Question 2

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

Question 3

What is the value of x?

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

Question 4

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

Question 5

Solve for x:

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

Exercise #6

Solve for y:

y2+4y+2=−2 y^2+4y+2=-2

Video Solution

Step-by-Step Solution

Let's solve the given equation:

y2+4y+2=−2 y^2+4y+2=-2

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

y2+4y+2=−2y2+4y+2+2=0y2+4y+4=0 y^2+4y+2=-2 \\ y^2+4y+2+2=0 \\ y^2+4y+4=0

Now we notice that the expression on the left side can be factored 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

We can do this using the fact that:

4=22 4=2^2

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

y2+4y+4=0↓y2+4y+22=0 y^2+4y+4=0 \\ \downarrow\\ \textcolor{red}{y}^2+4y+\textcolor{blue}{2}^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 we got in the last step:

y2+4y‾+22=0 \textcolor{red}{y}^2+\underline{4y}+\textcolor{blue}{2}^2=0

Notice that the terms y2,22 \textcolor{red}{y}^2,\hspace{6pt}\textcolor{blue}{2}^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):

(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 we can represent the expression on the left side of the equation as:

y2+4y‾+22=0↕?y2+2⋅y⋅2‾+22=0 \textcolor{red}{y}^2+\underline{4y}+\textcolor{blue}{2}^2 =0 \\ \updownarrow\text{?}\\ \textcolor{red}{y}^2+\underline{2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{2}}+\textcolor{blue}{2}^2 =0

And indeed it is true that:

2⋅y⋅2=4y 2\cdot y\cdot2=4y

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

y2+2⋅y⋅2+22=0↓(y+2)2=0 \textcolor{red}{y}^2+2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{2}+\textcolor{blue}{2}^2=0 \\ \downarrow\\ (\textcolor{red}{y}+\textcolor{blue}{2})^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:

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

Let's summarize the solution of the equation:

y2+4y+2=−2y2+4y+4=0↓y2+2⋅y⋅2+22=0↓(y+2)2=0↓y+2=0↓y=−2 y^2+4y+2=-2 \\ y^2+4y+4=0 \\ \downarrow\\ \textcolor{red}{y}^2+2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{2}+\textcolor{blue}{2}^2=0 \\ \downarrow\\ (\textcolor{red}{y}+\textcolor{blue}{2})^2=0 \\ \downarrow\\ y+2=0\\ \downarrow\\ \boxed{y=-2}

Therefore the correct answer is answer D.

Answer

y=−2 y=-2

Exercise #7

4x2=12x−9 4x^2=12x-9

Video Solution

Answer

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

Exercise #8

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 #9

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

Video Solution

Answer

x=0,−1,2 x=0,-1,2

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

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

Question 3

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

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+x2−12x=0 x^3+x^2-12x=0

Video Solution

Answer

x=0,3,−4 x=0,3,-4

Exercise #13

x3−7x2+6x=0 x^3-7x^2+6x=0

Video Solution

Answer

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

Topics learned in later sections

  1. Methods for Solving a Quadratic Function
  2. Solution by extracting a root
  3. Completing the square in a quadratic equation