Question

1. choose efficient methods what is the completely factored form of the expression $16x^{4}-y^{4}$? describe the method(s) of factoring you used.
2. error analysis describe and correct the error a student made in factoring $x^{2}-36$.

use the perfect-square trinomial pattern to factor $x^{2}-36$ because both terms are perfect squares.
$x^{2}-36=(x-6)(x-6)$

Explanation:

Step1: Factor as difference of squares

$16x^4 - y^4 = (4x^2)^2 - (y^2)^2 = (4x^2 + y^2)(4x^2 - y^2)$

Step2: Factor remaining difference of squares

$4x^2 - y^2 = (2x)^2 - y^2 = (2x + y)(2x - y)$

Step3: Combine all factors

$16x^4 - y^4 = (4x^2 + y^2)(2x + y)(2x - y)$
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Step4: Identify error in problem 20

The student misapplied the perfect-square trinomial pattern to a difference of squares, and incorrectly factored $x^2-36$ as $(x-6)(x-6)$ instead of using the difference of squares rule.

Step5: Correct the factoring

$x^2 - 36 = x^2 - 6^2 = (x+6)(x-6)$

Answer:

1. The completely factored form of $16x^4 - y^4$ is $\boldsymbol{(4x^2 + y^2)(2x + y)(2x - y)}$. The method used is repeated application of the difference of squares factoring rule ($a^2 - b^2=(a+b)(a-b)$): first factor $16x^4 - y^4$ as $(4x^2 + y^2)(4x^2 - y^2)$, then factor $4x^2 - y^2$ further into $(2x + y)(2x - y)$.
2. Error: The student confused a difference of squares with a perfect-square trinomial. A perfect-square trinomial has three terms in the form $a^2\pm2ab+b^2$, but $x^2-36$ is a two-term difference of squares, not a trinomial. Additionally, the student incorrectly factored it as $(x-6)(x-6)$, which expands to $x^2-12x+36$, not $x^2-36$.

Corrected Factoring: $\boldsymbol{x^2 - 36=(x+6)(x-6)}$ using the difference of squares rule.