Question

11 multiple choice 1 point
$(2x - 3)(4x^2 + 6x + 9)$ is the correct factored form for which polynomial expression?
$8x^3 - 27$
$2x^3 - 3$
$8x^3 + 27$
$27x^3 - 8$
$27x^3 + 8$
clear my selection
12 fill in the blank 1 point
factor completely. $7x^3 - 3x^2 + 35x - 15$
fill in the values of the letters in the answer: $(ax^2 + b)(cx + d)$
a= type your answer...
b= type your answer...
c= type your answer...
d= type your answer...
13 multiple answer 1 point
solve. $3x^3 + 9x^2 - 12x = 36$

Explanation:

Question 11

Step1: Recall the formula for difference of cubes

The formula for the difference of cubes is \(a^3 - b^3=(a - b)(a^2+ab + b^2)\).

Step2: Identify \(a\) and \(b\) in the given factored form

In the expression \((2x - 3)(4x^2+6x + 9)\), we can see that \(a = 2x\) and \(b=3\) because \((a - b)(a^2+ab + b^2)=(2x-3)((2x)^2+(2x)\times3 + 3^2)=(2x - 3)(4x^2+6x + 9)\)

Step3: Apply the difference of cubes formula

Using the formula \(a^3 - b^3=(a - b)(a^2+ab + b^2)\), substitute \(a = 2x\) and \(b = 3\). Then \(a^3=(2x)^3=8x^3\) and \(b^3=3^3 = 27\). So \(a^3-b^3=8x^3-27\)

Step1: Group the terms

We have the polynomial \(7x^3-3x^2 + 35x-15\). Group the first two terms and the last two terms: \((7x^3-3x^2)+(35x - 15)\)

Step2: Factor out the GCF from each group

From the first group \(7x^3-3x^2\), the GCF is \(x^2\), so we get \(x^2(7x - 3)\). From the second group \(35x-15\), the GCF is \(5\), so we get \(5(7x - 3)\)

Step3: Factor out the common binomial factor

Now we have \(x^2(7x - 3)+5(7x - 3)\). The common binomial factor is \((7x - 3)\), so we factor it out: \((x^2 + 5)(7x-3)\)

Step4: Compare with the given form \((Ax^2 + B)(Cx+D)\)

Comparing \((x^2 + 5)(7x-3)\) with \((Ax^2 + B)(Cx+D)\), we get \(A = 1\), \(B=5\), \(C = 7\), \(D=- 3\)

Step1: Move all terms to one side

Given the equation \(3x^3+9x^2-12x = 36\), subtract \(36\) from both sides to get \(3x^3+9x^2-12x-36 = 0\)

Step2: Factor by grouping

Group the terms: \((3x^3+9x^2)+(-12x - 36)=0\)
Factor out the GCF from each group: \(3x^2(x + 3)-12(x + 3)=0\)

Step3: Factor out the common binomial factor

Factor out \((x + 3)\): \((x + 3)(3x^2-12)=0\)

Step4: Factor the quadratic expression

Factor \(3x^2-12\) by factoring out \(3\): \(3(x^2 - 4)=0\). And \(x^2-4\) is a difference of squares, so \(x^2-4=(x - 2)(x + 2)\). So the equation becomes \(3(x + 3)(x - 2)(x + 2)=0\)

Step5: Solve for \(x\)

Set each factor equal to zero:

• \(x+3=0\) gives \(x=-3\)
• \(x - 2=0\) gives \(x = 2\)
• \(x+2=0\) gives \(x=-2\)

Answer:

A. \(8x^3 - 27\)

Question 12