GMAT Math : Solving by Factoring

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Example Questions

Example Question #21 : Solving By Factoring

$x^{2}+4xy +4 y^{2} = 100$

$x^{2}-4xy +4 y^{2} = 64$

Which of the following is the set of possible values of ?

Possible Answers:

$\{-41,-9, 9, 41\}$

$\left \{ -8, -6, 6, 8\right \}$

$\{-9, -1, 1, 9 \}$

$\{-82,-18, 18 , 82\}$

$\left \{ -\frac{9}{2}, -\frac{1}{2}, \frac{1}{2}, \frac{9}{2} \right \}$

Correct answer:

$\{-9, -1, 1, 9 \}$

Explanation:

If you note that the expressions on the left sides of the equations are perfect square trinomials, you can rewrite the expressions as follows:

$x^{2}+4xy +4 y^{2} = 100$

$(x+2y)^{2} = 100$

Either x+2y = 10 or x+2y = -10

Similarly,

$x^{2}-4xy +4 y^{2} = 64$

$(x-2y)^{2} = 64$

Either x-2y= 8 or x-2y = -8

Four systems of equations can be set up here, and in each case, we can find by adding both sides:

x+2y = 10

$\underline{x-2y= 8}$

$2x\; \; \; \; \;=18$

$x\; \; \; \; \;= 9$

x+2y =- 10

$\underline{x-2y= -8}$

$2x\; \; \; \; \;=-18$

$x\; \; \; \; \;=-9$

x+2y = 10

$\underline{x-2y= -8}$

$2x\; \; \; \; \;= 2$

$x\; \; \; \; \;= 1$

x+2y =- 10

$\underline{x-2y= 8}$

$2x\; \; \; \; \;=-2$

$x\; \; \; \; \;=-1$

The set of possible values of is $\{-9, -1, 1, 9 \}$ .

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Example Question #21 : Solving By Factoring

$x^{2}+4xy +4 y^{2} = 100$

$x^{2}-4xy +4 y^{2} = 64$

Which of the following is the set of possible values of ?

Possible Answers:

$\{-9, -1, 1, 9 \}$

$\{-82,-18, 18 , 82\}$

$\left \{ -\frac{9}{2}, -\frac{1}{2}, \frac{1}{2}, \frac{9}{2} \right \}$

$\left \{ -8, -6, 6, 8\right \}$

$\{-41,-9, 9, 41\}$

Correct answer:

$\left \{ -\frac{9}{2}, -\frac{1}{2}, \frac{1}{2}, \frac{9}{2} \right \}$

Explanation:

If you note that the expressions on the left sides of the equations are perfect square trinomials, you can rewrite the expressions as follows:

$x^{2}+4xy +4 y^{2} = 100$

$(x+2y)^{2} = 100$

Either x+2y = 10 or x+2y = -10

Similarly,

$x^{2}-4xy +4 y^{2} = 64$

$(x-2y)^{2} = 64$

Either x-2y= 8 or x-2y = -8

Four systems of equations can be set up here, and in each case, we can find by subtracting both sides:

x+2y = 10

$\underline{x-2y= 8}$

4y = 2

$y = \frac{1}{2}$

x+2y =- 10

$\underline{x-2y= -8}$

4y = -2

$y =- \frac{1}{2}$

x+2y = 10

$\underline{x-2y= -8}$

4y = 18

$y = \frac{9}{2}$

x+2y =- 10

$\underline{x-2y= 8}$

4y = -18

$y =- \frac{9}{2}$

The set of possible values of is $\left \{ -\frac{9}{2}, -\frac{1}{2}, \frac{1}{2}, \frac{9}{2} \right \}$ .

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Example Question #21 : Solving By Factoring

Solve for by factoring:

x^2+7x+10=0

Possible Answers:

x=2,x=5

x=-1,x=-10

x=-10,x=7

x=7,x=10

x=-2,x=-5

Correct answer:

x=-2,x=-5

Explanation:

When we factor a polynomial, we are left with two factors of the form:

(x+a)(x+b)=0

For the given function, this means we must find two factors and whose product is 10 and whose sum is 7. Thinking about factors of 10, we can see that 1 and 10 cannot add up to 7 in any way. The only two factors left, then, are 2 and 5, which we can see have a product of 10 and a sum of 7. This gives us:

x^2+7x+10=0

(x+2)(x+5)=0

x=-2,x=-5

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Example Question #21 : Solving By Factoring

Find the roots of the following function:

f(x)=2x^2+6x-20

Possible Answers:

x=2,x=4

x=-3,x=10

x=-2,x=5

x=-10,x=3

x=-5,x=2

Correct answer:

x=-5,x=2

Explanation:

The roots of a function are the points at which it crosses the axis, so f(x) will be equal to zero at these points. This means we set the function equal to zero, and then factor it to solve for the values of the roots:

f(x)=2x^2+6x-20=0

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

Now that we've factored out a , we can see we need two factors whose product is and whose sum is . Thinking about the possible factors, we can see that and have a product of and a sum of . This gives us:

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

2(x+5)(x-2)=0

x=-5,x=2

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Example Question #431 : Algebra

Solve for .

(x-3)^2=2x^2+14

Possible Answers:

x=2,3

x=-1, -5

x=-2,-3

x=1, 5

Correct answer:

x=-1, -5

Explanation:

In order to solve this problem, we need to combine like terms and then factor:

(x-3)^2=2x^2+14

x^2-6x+9=2x^2+14

0=x^2+6x+5

0=(x+5)(x+1)

x=-1, -5

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Example Question #432 : Algebra

Solve for :

$4x^{3} - 28x^{2} + 48x = 0$

Possible Answers:

x=28,48

x=0,-3,-4

x=-3,-4

x=3,4

x=0,3,4

Correct answer:

x=0,3,4

Explanation:

To solve this we must factor:

The first step is to recognize that all of the terms on the equation's left side contain 4x. Therefore we can pull 4x out:

$4x(x^{2}-7x +12) = 0$

Then we factor the inside of the parentheses, realizing that only -3, and -4 add to form -7, and multiply to form 12:

4x(x - 3)(x - 4) = 0

Now we can see that, for the left side of the equation to equal zero, x can only equal 0, 3, or 4.

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Example Question #21 : Solving By Factoring

Solve the following by factoring:

2x^2-6x-8=0

Possible Answers:

x=4,-1

x=6,-2

x=-4,1

x=2,-6

Correct answer:

x=4,-1

Explanation:

To solve, divide out a and then factor.

2(x^2-3x-4)=0

x^2-3x-4=0

(x-4)(x+1)=0

x=4,-1

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Example Question #22 : Solving By Factoring

Simplify $\frac{3x^{2}-27}{x-3}$ .

Possible Answers:

3x+3

$\frac{3x+3}{x-3}$

3x+9

x+3

$\frac{3}{x-3}$

Correct answer:

3x+9

Explanation:

$\frac{3x^{2}-27}{x-3}=\frac{3(x^{2}-9)}{x-3}=\frac{3(x-3)(x+3)}{x-3}$

=3(x+3)=3x+9.

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GMAT Math : Solving by Factoring

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #21 : Solving By Factoring

Example Question #21 : Solving By Factoring

Example Question #21 : Solving By Factoring

Example Question #21 : Solving By Factoring

Example Question #431 : Algebra

Example Question #432 : Algebra

Example Question #21 : Solving By Factoring

Example Question #22 : Solving By Factoring

All GMAT Math Resources

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