GMAT Math : Algebra

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Example Questions

Example Question #411 : Algebra

Factor the expression completely:

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

Possible Answers:

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

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

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

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

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

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 #2 : Understanding Factoring

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

Where does this function cross the -axis?

Possible Answers:

x=5, 10

x=9, 18

x=0

x=3, 6

It never crosses the x axis.

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 #412 : Algebra

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 #413 : Algebra

Factor:

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

Possible Answers:

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

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

(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 #414 : Algebra

Factor completely:

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

Possible Answers:

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

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

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

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

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

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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Example Question #411 : Algebra

Factor:

$4x^{2}-12xy +9y^{2} - 64x^{2}y^{2}$

Possible Answers:

$\left ( 2x+3y +8xy \right ) \left ( 2x-3y -8xy \right )$

$\left ( 2x-3y +8xy \right ) \left ( 2x-3y -8xy \right )$

$\left ( 2x-3y -8xy \right )^{2}$

$\left ( 2x+3y +8xy \right ) \left ( 2x+3y -8xy \right )$

Correct answer:

$\left ( 2x-3y +8xy \right ) \left ( 2x-3y -8xy \right )$

Explanation:

$4x^{2}-12xy +9y^{2} - 64x^{2}y^{2}$ can be grouped as follows:

$\left (4x^{2}-12xy +9y^{2} \right ) - 64x^{2}y^{2}$

The first three terms form a perfect square trinomial, since $2\cdot \sqrt{4x^{2} } \cdot \sqrt { 9y^{2}} = 2 \cdot 2x \cdot 3y = 12xy$

$4x^{2}-12xy +9y^{2} = \left ( 2x - 3y\right )^{2}$ , so

$\left (4x^{2}-12xy +9y^{2} \right ) - 64x^{2}y^{2}$

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

$= \left ( 2x - 3y\right )^{2} - \left ( 8 xy \right ) ^{2}$

Now use the dfference of squares pattern:

$= \left [ \left ( 2x - 3y \right ) + 8 xy \right ] \left [ \left ( 2x - 3y \right ) - 8 xy \right ]$

$= \left ( 2x - 3y + 8 xy \right ) \left ( 2x - 3y - 8 xy \right )$

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Example Question #415 : Algebra

Solve for when f(x)=0 .

f(x)=x^2+14x+49

Possible Answers:

x=49

x=14

x=7

x=-7

Correct answer:

x=-7

Explanation:

f(x)=x^2+14x+49

0=x^2+14x+49

0=(x+7)(x+7)

x+7=0

x=-7

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Example Question #416 : Algebra

Solve for when f(x)=0 .

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

Possible Answers:

and x=-1

and

x=2 and

x=2 and x=-1

Correct answer:

and

Explanation:

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

0=x^2+x-2

0=(x+2)(x-1)

x+2=0 and x-1=0

and

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Example Question #417 : Algebra

Solve for :

x^2-8x-9=0

Possible Answers:

x=9 and x=1

x=-9 and x=1

x=9 and x=-1

x=-9 and x=-1

Correct answer:

x=9 and x=-1

Explanation:

x^2-8x-9=0

(x-9)(x+1)=0

x-9=0 and x+1=0

x=9 and x=-1

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GMAT Math : Algebra

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #411 : Algebra

Example Question #2 : Understanding Factoring

Example Question #412 : Algebra

Example Question #413 : Algebra

Example Question #414 : Algebra

Example Question #1 : Solving By Factoring

Example Question #411 : Algebra

Example Question #415 : Algebra

Example Question #416 : Algebra

Example Question #417 : Algebra

All GMAT Math Resources

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