Question

38 multiple choice 1 point
factor completely.
$x^4 - x^2 - 12$
$\big(x^2 - 4\big)\big(x^2 + 3\big)$
$\big(x + 2\big)\big(x - 2\big)\big(x^2 - 3\big)$
$\big(x + 4\big)\big(x - 4\big)\big(x^2 + 3\big)$
$\big(x + 2\big)\big(x - 2\big)\big(x^2 + 3\big)$
$\big(x^2 + 4\big)\big(x^2 - 3\big)$
clear my selection
39 multiple answer 1 point
which are factors of $f(x) = x^3 - 2x^2 - 5x + 6$? choose all that apply.
x-3
x-1
none of these
x+2
x+3
x-2
x+1
40 multiple choice 1 point
factor completely. $x^3 + 64$
$\big(x - 4\big)\big(x^2 + 4x + 16\big)$
$\big(x + 4\big)^3$
$\big(x + 4\big)\big(x^2 - 4x + 16\big)$
$\big(x + 4\big)\big(x^2 + 4x + 16\big)$

Explanation:

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Question 38

Step1: Substitute $u=x^2$

$x^4 - x^2 - 12 = u^2 - u - 12$

Step2: Factor quadratic in $u$

$u^2 - u - 12 = (u-4)(u+3)$

Step3: Substitute back $u=x^2$

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

Step4: Factor difference of squares

$x^2-4=(x+2)(x-2)$, so final form: $(x + 2)(x - 2)(x^2 + 3)$

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Question 39

Step1: Test $x=3$ in $f(x)$

$f(3)=3^3-2(3)^2-5(3)+6=27-18-15+6=0$, so $x-3$ is a factor.

Step2: Test $x=1$ in $f(x)$

$f(1)=1^3-2(1)^2-5(1)+6=1-2-5+6=0$, so $x-1$ is a factor.

Step3: Test $x=-2$ in $f(x)$

$f(-2)=(-2)^3-2(-2)^2-5(-2)+6=-8-8+10+6=0$, so $x+2$ is a factor.

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Question 40

Step1: Recognize sum of cubes

$x^3+64=x^3+4^3$

Step2: Apply sum of cubes formula

$a^3+b^3=(a+b)(a^2-ab+b^2)$, where $a=x, b=4$
$x^3+4^3=(x+4)(x^2-4x+16)$

Answer:

1. Question 38: $(x + 2)(x - 2)(x^2 + 3)$
2. Question 39: x-3, x-1, x+2
3. Question 40: $(x + 4)(x^2 - 4x + 16)$