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ACT Math : How to factor an equation

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Example Questions

Example Question #1 : Factoring Equations

Solve 8x² – 2x – 15 = 0

Possible Answers:

x = 3/2 or 5/4

x = 3/2 or -5/4

x = -3/2 or 5/4

x = -3/2 or -5/4

Correct answer:

x = 3/2 or -5/4

Explanation:

The equation is in standard form, so a = 8, b = -2, and c = -15. We are looking for two factors that multiply to ac or -120 and add to b or -2. The two factors are -12 and 10.

So you get (2x -3)(4x +5) = 0. Set each factor equal to zero and solve.

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Example Question #1 : Factoring Equations

If (x²+ 2) / 2 = (x² - 6x - 1) / 5, then what is the value of x?

Possible Answers:

-2

-3

Correct answer:

-2

Explanation:

(x²+ 2) / 2 = (x² - 6x - 1) / 5. We first cross-multiply to get rid of the denominators on both sides.

5(x² + 2) = 2(x² - 6x - 1)

5x² + 10 = 2x² - 12x - 2 (Subtract 2x², and add 12x and 2 to both sides.)

3x² + 12x + 12 = 0 (Factor out 3 from the left side of the equation.)

3(x² + 4x + 4) = 0 (Factor the equation, knowing that 2 + 2 = 4 and 2*2 = 4.)

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

x + 2 = 0

x = -2

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Example Question #2 : Factoring Equations

Which of the following is a factor of the polynomial x² – 6x + 5?

Possible Answers:

x – 5

x + 2

x – 6

x – 8

x + 1

Correct answer:

x – 5

Explanation:

Factor the polynomial by choosing values that when FOIL'ed will add to equal the middle coefficient, 3, and multiply to equal the constant, 1.

x² – 6x + 5 = (x – 1)(x – 5)

Because only (x – 5) is one of the choices listed, we choose it.

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Example Question #3 : Factoring Equations

7 times a number is 30 less than that same number squared. What is one possible value of the number?

Possible Answers:

$0$

$-3$

$1$

$-10$

$3$

Correct answer:

$-3$

Explanation:

$\small 7x+30=x^{2}$

$\small x^{2}-7x-30=0$

$\small (x-10)(x+3)=0$

Either:

$\small x-10=0$

$\small x=10$

or:

$\small x+3=0$

$\small x=-3$

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Example Question #4 : Factoring Equations

Which of the following is equivalent to $2x^{2}\left ( xy^{2}+5x^{2}y^{2} \right )$ ?

Possible Answers:

$2x^{3}y^{2}+x^{4}y^{2}$

$2xy^{2}+x^{4}y^{2}$

$2x^{3}y^{2}+10x^{4}y$

$2x^{3}y^{2}+10x^{4}y^{2}$

$x^{3}y^{2}+10x^{4}y^{2}$

Correct answer:

$2x^{3}y^{2}+10x^{4}y^{2}$

Explanation:

The answer is $2x^{3}y^{2}+10x^{4}y^{2}$ ^.

To determine the answer, $2x^{2}$ must be distrbuted,

$(2x^{2}*xy^{2}) +(2x^{2}*5x^{2}y^{2})$ . After multiplying the terms, the expression simplifies to $2x^{3}y^{2}+10x^{4}y^{2}$ .

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Example Question #5 : Factoring Equations

For what value of b is the equation b² + 6b + 9 = 0 true?

Possible Answers:

–3

Correct answer:

–3

Explanation:

Factoring leads to (b+3)(b+3)=0. Therefore, solving for b leads to -3.

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Example Question #6 : Factoring Equations

What is the solution to:

$\frac{x^2-6x+8}{x-2}=0$

Possible Answers:

Correct answer:

Explanation:

First you want to factor the numerator from x²– 6x + 8 to (x – 4)(x – 2)

Input the denominator (x – 4)(x – 2)/(x – 2) = (x – 4) = 0, so x = 4.

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Example Question #7 : Factoring Equations

What is the value of where:

3x+5=6-2x

Possible Answers:

$-\frac{1}{5}$

$\frac{1}{5}$

Correct answer:

$\frac{1}{5}$

Explanation:

The question asks us to find the value of , because it is in a closed equation, we can simply put all of the whole numbers on one side of the equation, and all of the containing numbers on the other side.

We utilize opposite operations to both sides by adding to each side of the equation and get 5x+5=6

Next, we subtract from both sides, yielding

5x=1

Then we divide both sides by to get rid of that on

$x=\frac{1}{5}$

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Example Question #8 : Factoring Equations

Factor the following equation:

${x^3-2x^2+x}$

Possible Answers:

$(x-1)^{3}$

x(x-1)(x-1)

$x^{2}(x-1)$

$x^{2}(x-1)^{2}$

$(x-1)^{2}$

Correct answer:

x(x-1)(x-1)

Explanation:

First we factor out an x then we can factor the (x-1)(x-1)

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Example Question #203 : Algebra

Which of the following equations is NOT equivalent to the following equation?

$4y^{2}=169x^{2}-81$

Possible Answers:

$y=\sqrt{\frac{338x^{2}-162}{8}}$

$\frac{(2y)^{2}}{13x+9}=27x-9-14x$

$6y^{2}=\frac{3\times (13x+9)\times (13x-9)}{2}$

$y^{2}=(\frac{13x-9}{2})^{2}$

$(2y)^{2}=(13x+9)(13x-9)$

Correct answer:

$y^{2}=(\frac{13x-9}{2})^{2}$

Explanation:

The equation presented in the problem is:

$4y^{2}=169x^{2}-81$

We know that:

$169x^{2}-81=(13x)^{2}-9^{2}=(13x+9)(13x-9)$

Therefore we can see that the answer choice $(2y)^{2}=(13x+9)(13x-9)$ is equivalent to $4y^{2}=169x^{2}-81$ .

$\frac{(2y)^{2}}{13x+9}=27x-9-14x$ is equivalent to $4y^{2}=169x^{2}-81$ . You can see this by first combining like terms on the right side of the equation:

$\frac{(2y)^{2}}{13x+9}=13x-9$

Multiplying everything by 13x+9 , we get back to:

$(2y)^{2}=(13x+9)(13x-9)$

We know from our previous work that this is equivalent to $4y^{2}=169x^{2}-81$ .

$6y^{2}=\frac{3\times (13x+9)\times (13x-9)}{2}$ is also equivalent $4y^{2}=169x^{2}-81$ since both sides were just multiplied by $\frac{3}{2}$ . Dividing both sides by $\frac{3}{2}$ , we also get back to:

$(2y)^{2}=(13x+9)(13x-9)$ .

We know from our previous work that this is equivalent to $4y^{2}=169x^{2}-81$ .

$y=\sqrt{\frac{338x^{2}-162}{8}}$ is also equivalent to $4y^{2}=169x^{2}-81$ since

$y^{2}=(\sqrt{\frac{338x^{2}-162}{8}})^{2}=\frac{338x^{2}-162}{8}$

$4y^{2}=\frac{338x^{2}-162}{2}=169x^{2}-81$

Only $y^{2}=(\frac{13x-9}{2})^{2}$ is NOT equivalent to $4y^{2}=169x^{2}-81$

because

$y^{2}=(\frac{13x-9}{2})^{2}=\frac{169x^{2}-234x-81}{4}$

$4y^{2}=169x^{2}-81-234x \neq 4y^{2}=169x^{2}-81$

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ACT Math : How to factor an equation

Study concepts, example questions & explanations for ACT Math

All ACT Math Resources

Example Questions

Example Question #1 : Factoring Equations

Example Question #1 : Factoring Equations

Example Question #2 : Factoring Equations

Example Question #3 : Factoring Equations

Example Question #4 : Factoring Equations

Example Question #5 : Factoring Equations

Example Question #6 : Factoring Equations

Example Question #7 : Factoring Equations

Example Question #8 : Factoring Equations

Example Question #203 : Algebra

All ACT Math Resources

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