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Algebra II : Intermediate Single-Variable Algebra

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Example Questions

Example Question #12 : Factoring Polynomials

Factor the following expression:

x^2 -10x - 24

Possible Answers:

(x-12)(x+2)

(x+3)(x-8)

(x-12)(x-2)

(x-6)(x+4)

(x-24)(x-1)

Correct answer:

(x-12)(x+2)

Explanation:

x^2 -10x - 24

To factor, we are looking for two terms that multiply to give and add together to get .

Possible factors of :

$(-1,24),\ (1,-24),\ (-2,12),\ (2,-12),\ (-3,8),\ (3,-8),\ (-4,6),\ (4,-6)$

Based on these options, it is clear our factors are and .

$(2)(-12)=-24\ \text{and}\ 2+(-12)=-10$

Our final answer will be:

(x-12)(x+2)

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Example Question #61 : Intermediate Single Variable Algebra

Factor the following expression:

x^2 +7x - 8

Possible Answers:

(x-4)(x-2)

(x-8)(x+1)

(x+4)(x-4)

(x+4)(x-2)

(x+8)(x-1)

Correct answer:

(x+8)(x-1)

Explanation:

x^2 +7x - 8

To factor, we are looking for two terms that multiply to give and add together to get .

Possible factors of :

$(-1,8),\ (1,-8),\ (-2,4),\ (2,-4)$

Based on these options, it is clear our factors are and .

$(-1)(8)=-8\ \text{and}\ (-1)+8=7$

Our final answer will be:

(x+8)(x-1)

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Example Question #14 : Factoring Polynomials

Factor the following expression:

x^2 - 11x + 10

Possible Answers:

(x-5)(x+2)

(x-10)(x-1)

(x-5)(x-5)

(x-2)(x-5)

(x-10)(x+1)

Correct answer:

(x-10)(x-1)

Explanation:

x^2 - 11x + 10

To factor, we are looking for two terms that multiply to give and add together to get .

Possible factors of :

$(1,10),\ (-1,-10),\ (2,5),\ (-2,-5)$

Based on these options, it is clear our factors are and .

$(-1)(-10)=10\ \text{and}\ (-1)+(-10)=-11$

Our final answer will be:

(x-10)(x-1)

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Example Question #61 : Intermediate Single Variable Algebra

Factor the following expression:

x^2 + 20x + 96

Possible Answers:

(x + 2)(x-48)

(x + 12)(x+8)

(x+24)(x+4)

(x + 6)(x+16)

(x - 12)(x-8)

Correct answer:

(x + 12)(x+8)

Explanation:

x^2 + 20x + 96

To factor, we are looking for two terms that multiply to give and add together to get . There are numerous factors of , so we will only list a few.

Possible factors of :

$(3,32),\ (-3,-32),\ (4,24),\ (-4,-24),\ (6,16),\ (8,12)$

Based on these options, it is clear our factors are and .

$(8)(12)=96\ \text{and}\ 8+12=20$

Our final answer will be:

(x + 12)(x+8)

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Example Question #13 : Factoring Polynomials

Factor the following expression:

x^2 -38x -600

Possible Answers:

(x-60)(x+10)

(x-50)(x+12)

(x-300)(x+2)

(x-50)(x-12)

(x-100)(x+6)

Correct answer:

(x-50)(x+12)

Explanation:

x^2 -38x -600

To factor, we are looking for two terms that multiply to give and add together to get . There are numerous factors of , so we will only list a few.

Possible factors of :

$(2,-300),\ (3,-200),\ (5,-120),\ (6,-100),\ (10,-60),\ (12,-50),\ (15,-40)$

Based on these options, it is clear our factors are and .

$(12)(-50)=-600\ \text{and}\ 12+(-50)=-38$

Our final answer will be:

(x-50)(x+12)

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Example Question #14 : Factoring Polynomials

Factor the following expression:

x^2 - 26x + 105

Possible Answers:

(x - 5)(x-21)

(x + 7)(x-15)

(x - 5)(x+21)

(x - 7)(x-15)

(x + 7)(x-21)

Correct answer:

(x - 5)(x-21)

Explanation:

x^2 - 26x + 105

To factor, we are looking for two terms that multiply to give 105 and add together to get .

Possible factors of 105 :

$(3,35),\ (-3,-35),\ (5,21),\ (-5,-21),\ (7,15),\ (-7,-15)$

Based on these options, it is clear our factors are and .

$(-5)(-21)=105\ \text{and}\ (-5)+(-21)=-26$

Our final answer will be:

(x - 5)(x-21)

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Example Question #11 : Factoring Polynomials

Simplify:

$\frac{x^2+6x+9}{x+3}$

Possible Answers:

x+15

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

x+6x+3

x+3

Correct answer:

x+3

Explanation:

Factor the numerator:

$x^2+6x+9\rightarrow (x+3)(x+3)$

Simplify the fraction:

$\frac{(x+3)(x+3)}{(x+3)}\rightarrow (x+3)$

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Example Question #21 : Factoring Polynomials

Factor:

x^3-6x^2+9x

Possible Answers:

x(x+3)(x-3)

x(x-3)(x-3)

(x-9)(x+1)

x(x^2-6x+9)

x^2(x-3)

Correct answer:

x(x-3)(x-3)

Explanation:

Factor:

x^3-6x^2+9x

Step 1: Factor out

x(x^2-6x+9)

Step 2: Factor the polynomial

x(x-3)(x-3)

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Example Question #22 : Factoring Polynomials

Factor:

x^2+5x-14

Possible Answers:

(x-2)(x+7)

(x+7)(x-7)

(x+5)(x+2)

(x+14)(x-1)

(x-8)(x+8)

Correct answer:

(x-2)(x+7)

Explanation:

Factor:

x^2+5x-14

When factoring a polynomial ax^2+bx+c , the product of the coefficients must be , the sum of the factors must be , and the product of the factors must be .

For the above equation, a=1 , b=5 , and c=-14 .

Set up the factor equation:

$(x+{\color{Red} ?})(x+{\color{Red} ?})$

Becauase is negative, one of the factors must be negative as well. Because is positive, this means the larger factor is positive as well.

Two numbers that meet these requirements are and . Their product is , and their sum is .

(x-2)(x+7)

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Example Question #22 : Factoring Polynomials

Expand:

(x+3)(x-9)

Possible Answers:

x^2+6x-3

x^2-6x-27

x^2+3x+9

x^2-12x-27

x^2+6x-9

Correct answer:

x^2-6x-27

Explanation:

Follow the FOIL rule when multiplying - first, outside, inside, last.

First: $x\cdot x=x^2$

Oustide: $x\cdot -9=-9x$

Inside: $3\cdot x=3x$

Last: $3\cdot -9=-27$

Add all of these together:

x^2-9x+3x-27

Combine like terms:

${\color{Red} x^2-6x-27}$

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Algebra II : Intermediate Single-Variable Algebra

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #12 : Factoring Polynomials

Example Question #61 : Intermediate Single Variable Algebra

Example Question #14 : Factoring Polynomials

Example Question #61 : Intermediate Single Variable Algebra

Example Question #13 : Factoring Polynomials

Example Question #14 : Factoring Polynomials

Example Question #11 : Factoring Polynomials

Example Question #21 : Factoring Polynomials

Example Question #22 : Factoring Polynomials

Example Question #22 : Factoring Polynomials

All Algebra II Resources

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