Math  /  Algebra

QuestionIn Exercises 12-14, factor the polynomial completely. xx  12. 36p425(6pz5)(6p2+5)(6p2+5)()\begin{array}{l} \text { 12. } 36 p^{4}-25 \\ \left(6 p^{z-5}\right)\left(6 p^{2}+5\right) \\ \left(6 p^{2}+5\right)(\quad) \end{array}
13. n4+11n2+28n^{4}+11 n^{2}+28  *14. 4y2164(y24)(y2+4)(y2+4)(y2)(y+2)\begin{array}{l} \text { *14. } \begin{array}{l} \\ 4 \\ y^{2}-16 \\ 4 \\ \left(y^{2}-4\right)\left(y^{2}+4\right) \\ \left(y^{2}+4\right)(y-2)(y+2) \end{array} \end{array}

Studdy Solution

STEP 1

1. We are asked to factor the polynomial completely.
2. The polynomial given is n4+11n2+28 n^4 + 11n^2 + 28 .
3. We will use substitution to simplify the polynomial and then factor it.

STEP 2

1. Use substitution to simplify the polynomial.
2. Factor the resulting quadratic expression.
3. Substitute back to get the factors of the original polynomial.

STEP 3

Let u=n2 u = n^2 . Then the polynomial becomes:
u2+11u+28 u^2 + 11u + 28
This substitution simplifies the polynomial into a quadratic form.

STEP 4

Factor the quadratic expression u2+11u+28 u^2 + 11u + 28 .
We need two numbers that multiply to 28 and add to 11. These numbers are 7 and 4.
u2+11u+28=(u+7)(u+4) u^2 + 11u + 28 = (u + 7)(u + 4)

STEP 5

Substitute back u=n2 u = n^2 into the factored expression:
(n2+7)(n2+4) (n^2 + 7)(n^2 + 4)
This gives us the factors of the original polynomial.
The completely factored form of the polynomial n4+11n2+28 n^4 + 11n^2 + 28 is:
(n2+7)(n2+4) (n^2 + 7)(n^2 + 4)

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