QuestionWhich expression is equivalent to $162 x^{4}-144 x^{2}+32$ ? Select all that apply.
A. $2\left(81 x^{2}-72 x+16\right)$ B. $2\left(81 x^{2}+4\right)\left(81 x^{2}+4\right)$ C. $2\left(81 x^{2}-4\right)\left(81 x^{2}+4\right)$ D. $2\left(9 x^{2}-4\right)\left(9 x^{2}-4\right)$ E. $2\left(9 x^{2}+4\right)\left(9 x^{2}+4\right)$
5. F. $2(3 x+2)^{2}(3 x-2)^{2}$
The expression $x^{4}-81$ can be rewritten in this form where $a, b$ , and $c$ are real numbers: $\left(x^{2}+a\right)(x+b)(x+c)$
What are the values of $a, b$ , and $c$ ? Enter your answers in the boxes. $\begin{array}{l} a=\square \\ b=\square \\ c=\square \end{array}$ 6.
The expression $x^{2}(x-y)^{3}-y^{2}(x-y)^{3}$ can be written in the form $(x-y)^{a}(x+y)$ , where $a$ is a constant. What is the value of $a$ ? Enter your answer in the box. $a=$ $\square$
Write the expression $x-x y^{2}$ as the product of the greatest common factor and a binomial.

Studdy Solution

STEP 1

1. We need to factor the expression $162 x^{4}-144 x^{2}+32$ to find equivalent expressions.
2. We will use algebraic identities and factorization techniques.
3. We need to solve multiple parts of the problem, each requiring different approaches.

STEP 2

1. Factor the expression $162 x^{4}-144 x^{2}+32$ .
2. Solve the expression $x^{4}-81$ in the given form.
3. Simplify the expression $x^{2}(x-y)^{3}-y^{2}(x-y)^{3}$ .
4. Factor the expression $x-x y^{2}$ .

STEP 3

Factor the expression $162 x^{4}-144 x^{2}+32$ .
First, factor out the greatest common factor, which is 2:
$162 x^{4}-144 x^{2}+32 = 2(81 x^{4} - 72 x^{2} + 16)$
Now, recognize this as a quadratic in terms of $x^2$ :
$81(x^2)^2 - 72(x^2) + 16$
This can be factored further using the quadratic formula or by recognizing it as a perfect square trinomial:
$81(x^2)^2 - 72(x^2) + 16 = (9x^2 - 4)^2$
Thus, the expression becomes:
$2(9x^2 - 4)(9x^2 - 4)$

STEP 4

Solve the expression $x^{4}-81$ in the given form $(x^{2}+a)(x+b)(x+c)$ .
Recognize $x^4 - 81$ as a difference of squares:
$x^4 - 81 = (x^2 - 9)(x^2 + 9)$
Further factor $x^2 - 9$ as:
$x^2 - 9 = (x - 3)(x + 3)$
Thus, the expression becomes:
$(x - 3)(x + 3)(x^2 + 9)$
Comparing with $(x^{2}+a)(x+b)(x+c)$ , we have:
$a = 9, \, b = -3, \, c = 3$

STEP 5

Simplify the expression $x^{2}(x-y)^{3}-y^{2}(x-y)^{3}$ .
Factor out the common term $(x-y)^{3}$ :
$x^{2}(x-y)^{3} - y^{2}(x-y)^{3} = (x-y)^{3}(x^{2} - y^{2})$
Recognize $x^{2} - y^{2}$ as a difference of squares:
$x^{2} - y^{2} = (x+y)(x-y)$
Thus, the expression becomes:
$(x-y)^{3}(x+y)(x-y) = (x-y)^{4}(x+y)$
So, the value of $a$ is:
$a = 4$

STEP 6

Factor the expression $x-x y^{2}$ .
Factor out the greatest common factor, $x$ :
$x - xy^2 = x(1 - y^2)$
Recognize $1 - y^2$ as a difference of squares:
$1 - y^2 = (1 - y)(1 + y)$
Thus, the expression becomes:
$x(1-y)(1+y)$
The equivalent expressions for $162 x^{4}-144 x^{2}+32$ are: - C. $2(81 x^{2}-4)(81 x^{2}+4)$ - D. $2(9 x^{2}-4)(9 x^{2}-4)$
The values for $a, b, c$ are: $a = 9$ $b = -3$ $c = 3$
The value of $a$ in the expression $(x-y)^{a}(x+y)$ is: $a = 4$
The expression $x-x y^{2}$ as the product of the greatest common factor and a binomial is: $x(1-y^2) = x(1-y)(1+y)$

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