Question

12.) use multiple methods, factor the polynomial completely $27x^{2}z + 36xz + 12z$
answer:_____________________________
13.) use multiple methods, factor the polynomial completely $5x^{5}-80x$
answer:_____________________________
14.) using multiple methods factor the polynomial completely.
$2x^{5}+6x^{4}-32x - 96$
answer:_____________________________
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algebra 2a unit 3 module 5

Explanation:

Problem 12: Factor \( 27x^2z + 36xz + 12z \) completely

Step 1: Identify the GCF

Find the greatest common factor (GCF) of the terms \( 27x^2z \), \( 36xz \), and \( 12z \). The GCF of the coefficients \( 27 \), \( 36 \), and \( 12 \) is \( 3 \), and the common variable factor is \( z \). So the GCF is \( 3z \).
\[
27x^2z + 36xz + 12z = 3z(9x^2 + 12x + 4)
\]

Step 2: Factor the quadratic

Factor the quadratic \( 9x^2 + 12x + 4 \). Notice that it is a perfect square trinomial: \( (ax + b)^2 = a^2x^2 + 2abx + b^2 \). Here, \( a^2 = 9 \) (so \( a = 3 \)) and \( b^2 = 4 \) (so \( b = 2 \)), and \( 2ab = 2(3)(2) = 12 \), which matches the middle term.
\[
9x^2 + 12x + 4 = (3x + 2)^2
\]

Step 3: Combine the factors

Substitute the factored quadratic back into the expression.
\[
27x^2z + 36xz + 12z = 3z(3x + 2)^2
\]

Step 1: Identify the GCF

Find the GCF of \( 5x^5 \) and \( -80x \). The GCF of the coefficients \( 5 \) and \( 80 \) is \( 5 \), and the common variable factor is \( x \). So the GCF is \( 5x \).
\[
5x^5 - 80x = 5x(x^4 - 16)
\]

Step 2: Factor the difference of squares

The expression \( x^4 - 16 \) is a difference of squares: \( a^2 - b^2 = (a - b)(a + b) \), where \( a = x^2 \) and \( b = 4 \).
\[
x^4 - 16 = (x^2 - 4)(x^2 + 4)
\]

Step 3: Factor the remaining difference of squares

The term \( x^2 - 4 \) is also a difference of squares: \( a = x \) and \( b = 2 \).
\[
x^2 - 4 = (x - 2)(x + 2)
\]

Step 4: Combine the factors

Substitute the factored terms back into the expression.
\[
5x^5 - 80x = 5x(x - 2)(x + 2)(x^2 + 4)
\]

Step 1: Group the terms

Group the first two terms and the last two terms:
\[
(2x^5 + 6x^4) + (-32x - 96)
\]

Step 2: Factor out the GCF from each group

For the first group, the GCF of \( 2x^5 \) and \( 6x^4 \) is \( 2x^4 \). For the second group, the GCF of \( -32x \) and \( -96 \) is \( -32 \) (or \( 32 \), but factoring out \( -32 \) makes the binomial inside match the first group's binomial).
\[
2x^4(x + 3) - 32(x + 3)
\]

Step 3: Factor out the common binomial factor

The common binomial factor is \( (x + 3) \).
\[
(2x^4 - 32)(x + 3)
\]

Step 4: Factor the GCF from the first binomial

The GCF of \( 2x^4 \) and \( -32 \) is \( 2 \).
\[
2(x^4 - 16)(x + 3)
\]

Step 5: Factor the difference of squares

The term \( x^4 - 16 \) is a difference of squares: \( a = x^2 \) and \( b = 4 \).
\[
x^4 - 16 = (x^2 - 4)(x^2 + 4)
\]

Step 6: Factor the remaining difference of squares

The term \( x^2 - 4 \) is a difference of squares: \( a = x \) and \( b = 2 \).
\[
x^2 - 4 = (x - 2)(x + 2)
\]

Step 7: Combine the factors

Substitute the factored terms back into the expression.
\[
2x^5 + 6x^4 - 32x - 96 = 2(x - 2)(x + 2)(x^2 + 4)(x + 3)
\]

Answer:

\( 3z(3x + 2)^2 \)

Problem 13: Factor \( 5x^5 - 80x \) completely