GMAT Math : Understanding factoring

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Example Questions

Example Question #1 : Understanding Factoring

Factor $4y^{2}+4y-15$ .

Possible Answers:

$\dpi{100} \small (2y+3)(2y-5)$

$\dpi{100} \small (y-3)(y+5)$

$\dpi{100} \small (2y-5)(2y+5)$

$\dpi{100} \small (2y-3)(2y+3)$

$\dpi{100} \small (2y-3)(2y+5)$

Correct answer:

$\dpi{100} \small (2y-3)(2y+5)$

Explanation:

To factor this, we need two numbers that multiply to $\dpi{100} \small 4\times -15=60$ and sum to $\dpi{100} \small 4$ . The numbers $\dpi{100} \small -6$ and $\dpi{100} \small 10$ work.

$4y^{2}+4y-15 = 4y^{2} - 6y + 10y - 15 = 2y(2y-3) + 5(2y-3) = (2y-3)(2y+5)$

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Example Question #2 : Understanding Factoring

Factor $\frac{y-x}{x^{2}-y^{2}}$ .

Possible Answers:

$\frac{1}{x^{2}-y^{2}}$

$\frac{1}{x+y}$

$\frac{y-x}{x-y}$

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

$\frac{-1}{x+y}$

Correct answer:

$\frac{-1}{x+y}$

Explanation:

${x^{2}-y^{2}}$ is a difference of squares. The difference of squares formula is

${a^{2}-b^{2}} = (a-b)(a+b)$

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

Then, $\frac{y-x}{x^{2}-y^{2}} = \frac{y-x}{(x+y)(x-y)} = \frac{-(x-y)}{(x+y)(x-y)} = \frac{-1}{x+y}$ .

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Example Question #3 : Understanding Factoring

Solve $x^{2}-6x+5>0$ .

Possible Answers:

$x>5\ or\ x<1$

x>1

x<-1

x>5

1<x<5

Correct answer:

$x>5\ or\ x<1$

Explanation:

Let's factor the expression: $x^{2}-6x+5 = (x-1)(x-5)$ .

We need to look at the behavior of the function to the left and right of 1 and 5. To the left of x=1 ,

$x^{2}-6x+5>0$

You can check this by plugging in any value smaller than 1. For example, if x=0 ,

(x-1)(x-5)=5 ,

which is greater than 0.

When takes values in between 1 and 5, $x^{2}-6x+5 <0$ . Again we can check this by plugging in a number between 1 and 5.

x=2,(x-1)(x-5)=-3 , which is less than 0, so no numbers between 1 and 5 satisfy the inequality.

When takes values greater than 5, $x^{2}-6x+5>0$ .

To check, let's try x=6 . Then:

(x-1)(x-5)=5

so numbers greater than 5 also satisfy the inequality.

Therefore $x>5\ or\ x<1$ .

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Example Question #2 : Understanding Factoring

Solve $x^{2}+7x-8 <0$ .

Possible Answers:

x>-8

x<1

-8<x<1

$x<-8\ or\ x>1$

$x<-1\ or\ x>8$

Correct answer:

-8<x<1

Explanation:

First let's factor: $x^{2}+7x-8=(x+8)(x-1)$

x < -8: Let's try -10. (-10 + 8)(-10 - 1) = 22, so values less than -8 don't satisfy the inequality.

-8 < x < 1: Let's try 0. (0 + 8)(0 - 1) = -8, so values in between -8 and 1 satisfy the inequality.

x > 1: Let's try 2. (2 + 8)(2 - 1) = 10, so values greater than 1 don't satisfy the inequality.

Therefore the answer is -8 < x < 1.

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Example Question #3 : Understanding Factoring

Factor the expression completely:

$6,561x^{8}-256$

Possible Answers:

$(81x^{4}+ 16)\left (9x^{2} + 4)(3x + 2)^{2}$

$(81x^{4}+ 16)\left (9x^{2} + 4)(3x + 2)(3x - 2)$

$(81x^{4}+ 16)^{2}$

$(3x + 2)^{4}(3x - 2)^{4}$

$(9x^{2} + 4) ^{2}(9x^{2} - 4)^{2}$

Correct answer:

$(81x^{4}+ 16)\left (9x^{2} + 4)(3x + 2)(3x - 2)$

Explanation:

This expression can be rewritten:

$6,561x^{8}-256$

$(81x^{4})^{2}- 16^{2}$

As the difference of squares, this can be factored as follows:

$(81x^{4}+ 16)(81x^{4}- 16)$

As the sum of squares with relatively prime terms, the first factor is a prime polynomial. The second factor can be rewritten as the difference of two squares and factored:

$(81x^{4}+ 16)\left [(\left (9x^{2} \right )^{2}- 4^{2}) \right ]$

$(81x^{4}+ 16)\left (9x^{2} + 4) (9x^{2} - 4)$

Similarly, the middle polynomial is prime; the third factor can be rewritten as the difference of two squares and factored:

$(81x^{4}+ 16)\left (9x^{2} + 4) \left [(3x^{2} )^{2}- 2^{2} \right ]$

$(81x^{4}+ 16)\left (9x^{2} + 4)(3x + 2)(3x - 2)$

This is as far was we can factor, so this is the complete factorization.

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Example Question #6 : Understanding Factoring

$x^{2} - 9 x + 18$

Where does this function cross the -axis?

Possible Answers:

x=0

It never crosses the x axis.

x=3, 6

x=9, 18

x=5, 10

Correct answer:

x=3, 6

Explanation:

Factor the equation and set it equal to zero. (x-6)*(x-3)=0 . So the funtion will cross the -axis when

$x=3\ or\ x=6$

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Example Question #7 : Understanding Factoring

If $a^{2}-b^{2}=15$ , and a+b=5 , what is the value of a-b ?

Possible Answers:

Correct answer:

Explanation:

This questions tests the formula: a^2-b^2=(a+b)*(a-b) .

Therefore, we have 15=5(a-b) . So

$a-b=\frac{15}{5}=3$

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Example Question #1 : Understanding Factoring

Factor:

$64 - 4x^{2} + 20xy - 25y^{2}$

Possible Answers:

(8+ 2x-5y) (8- 2x-5y)

$(8+ 2x-5y) ^{2}$

(8- 2x-5y) (8- 2x+5y)

(8+ 2x-5y) (8- 2x+5y)

$(8- 2x+5y)^{2}$

Correct answer:

(8+ 2x-5y) (8- 2x+5y)

Explanation:

$64 - 4x^{2} + 20xy - 25y^{2}$ can be grouped as follows:

$64 - \left ( 4x^{2} - 20xy + 25y^{2} \right )$

$4x^{2} - 20xy + 25y^{2}$ is a perfect square trinomial, since

$2\cdot \sqrt{4x^{2}} \cdot \sqrt{25y^{2}} = 2 \cdot 2x \cdot 5y = 20xy$

$64 - \left ( 4x^{2} - 20xy + 25y^{2} \right )$

$= 8^{2} - \left ( 2x-5y \right )^{2}$

Now use the difference of squares pattern:

$= 8^{2} - \left ( 2x-5y \right )^{2}$

$=\left [ 8+ \left ( 2x-5y \right ) \right ]\left [ 8- \left ( 2x-5y \right ) \right ]$

= (8+ 2x-5y) (8- 2x+5y)

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Example Question #5 : Understanding Factoring

Factor completely:

$x^{5} + 3x^{4} + 2x^{3}+ 5x ^{2} + 15x + 10$

Possible Answers:

$\left (x^{3}+5x ^{2}+ 5 \right ) \left ( x+1\right ) \left ( x+2\right )$

$\left (x^{3}+5 \right ) \left ( x+1\right ) \left ( x+2\right )$

$\left (x^{3}-5 \right ) \left ( x-1\right ) \left ( x+2\right )$

$\left (x^{3}-5 \right ) \left ( x-2\right ) \left ( x+1\right )$

$\left (x^{3}+5x+ 5 \right ) \left ( x+1\right ) \left ( x+2\right )$

Correct answer:

$\left (x^{3}+5 \right ) \left ( x+1\right ) \left ( x+2\right )$

Explanation:

Group the first three terms and the last three terms, then factor out a GCF from each grouping:

$x^{5} + 3x^{4} + 2x^{3}+ 5x ^{2} + 15x + 10$

$=\left ( x^{5} + 3x^{4} + 2x^{3} \right )+\left ( 5x ^{2} + 15x + 10 \right )$

$=x^{3} \left ( x^{2} + 3x + 2 \right )+5 \left ( x^{2} + 3x + 2 \right )$

$=\left (x^{3}+5 \right )\left ( x^{2} + 3x + 2 \right )$

We try to factor $x^{3}+5$ as a sum of cubes; however, 5 is not a perfect cube, so the binomial is a prime.

To factor out $x^{2} + 3x + 2$ , we try to factor it into (x + ?)(x + ?) , replacing the question marks with two integers whose product is 2 and whose sum is 3. These integers are 1 and 2, so

$x^{2} + 3x + 2 = \left ( x+1\right ) \left ( x+2\right )$

The original polynomial has $\left (x^{3}+5 \right ) \left ( x+1\right ) \left ( x+2\right )$ as its factorization.

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Example Question #1 : Solving By Factoring

Factor completely: $x^{5} - 3x^{4} + x^{3}+ 8x ^{2} - 24x + 8$

Possible Answers:

$\left ( x+2\right ) \left ( x - 2 \right )^{2}( x^{2} - 3x +1)$

$\left ( x+2\right ) \left ( x^{2} - 2 x + 4 \right )( x^{2} - 3x +1)$

$\left ( x - 2 \right )^{3}( x^{2} - 3x -1)$

$\left ( x+2\right ) ^{2} \left ( x - 2 \right )( x^{2} - 3x -1)$

$\left ( x + 2 \right )^{3}( x^{2} - 3x +1)$

Correct answer:

$\left ( x+2\right ) \left ( x^{2} - 2 x + 4 \right )( x^{2} - 3x +1)$

Explanation:

Group the first three terms and the last three terms, then factor out a GCF from each grouping:

$x^{5} - 3x^{4} + x^{3}+ 8x ^{2} - 24x + 8$

$=( x^{5} - 3x^{4} + x^{3} ) +( 8x ^{2} - 24x + 8 )$

$=x^{3} ( x^{2} - 3x + 1 ) +8 ( x^{2} - 3x + 1 )$

$= \left ( x^{3} +8 \right )( x^{2} - 3x +1)$

$x^{3} +8$ is the sum of cubes and can be factored using this pattern:

$x^{3} +8$

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

$= \left ( x+2\right ) \left ( x^{2} - 2 \cdot x + 2^{2} \right )$

$= \left ( x+2\right ) \left ( x^{2} - 2 x + 4 \right )$

We try to factor out the quadratic trinomial as (x + ?)(x + ?) , replacing the question marks with integers whose product is 1 and whose sum is . These integers do not exist, so the trinomial is prime.

The factorization is therefore

$= \left ( x+2\right ) \left ( x^{2} - 2 x + 4 \right )( x^{2} - 3x +1)$

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GMAT Math : Understanding factoring

Study concepts, example questions & explanations for GMAT Math

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Example Question #2 : Understanding Factoring

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