How do I factor this expression completely?

Question

How do I factor this expression completely?

$4 x^{2} (x^{2} + 1) {(x + 2)}^{3} + 36 {(x^{2} + 1)}^{2} {(x + 2)}^{4}$ $4x^2(x^2+1)(x+2)^3+36(x^2+1)^2(x+2)^4$

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Answer 1 · 2016-06-30T16:42:31

$4 x^{2} (x^{2} + 1) {(x + 2)}^{2} + 36 {(x^{2} + 1)}^{2} {(x + 2)}^{4} = 4 (x^{2} + 1) {(x + 2)}^{2} (9 x^{4} + 36 x^{3} + 46 x^{2} + 36 x + 36)$ $4x^2(x^2+1)(x+2)^2+36(x^2+1)^2(x+2)^4=4(x^2+1)(x+2)^2(9x^4+36x^3+46x^2+36x+36)$

Explanation:

It is observed that $4 x^{2} (x^{2} + 1) {(x + 2)}^{2}$ $4x^2(x^2+1)(x+2)^2$ and $36 {(x^{2} + 1)}^{2} {(x + 2)}^{4}$ $36(x^2+1)^2(x+2)^4$ have common factors

$4$ $4$ , $(x^{2} + 1)$ $(x^2+1)$ and ${(x + 2)}^{2}$ $(x+2)^2$ .

Note that for $(x^{2} + 1)$ $(x^2+1)$ and $(x + 2)$ $(x+2)$ , we have selected the minimal power of these among two expressions. Similarly between $4$ $4$ and $36$ $36$ , $4$ $4$ is common factor.

Hence taking out these common factors, we get

$4 x^{2} (x^{2} + 1) {(x + 2)}^{2} + 36 {(x^{2} + 1)}^{2} {(x + 2)}^{4}$ $4x^2(x^2+1)(x+2)^2+36(x^2+1)^2(x+2)^4$

= $4 (x^{2} + 1) {(x + 2)}^{2} [x^{2} + 9 (x^{2} + 1) {(x + 2)}^{2}]$ $4(x^2+1)(x+2)^2[x^2+9(x^2+1)(x+2)^2]$

= $4 (x^{2} + 1) {(x + 2)}^{2} [x^{2} + 9 (x^{2} + 1) (x^{2} + 4 x + 4)]$ $4(x^2+1)(x+2)^2[x^2+9(x^2+1)(x^2+4x+4)]$

= $4 (x^{2} + 1) {(x + 2)}^{2} [x^{2} + 9 (x^{4} + 4 x^{3} + 4 x^{2} + x^{2} + 4 x + 4)]$ $4(x^2+1)(x+2)^2[x^2+9(x^4+4x^3+4x^2+x^2+4x+4)]$

= $4 (x^{2} + 1) {(x + 2)}^{2} (9 x^{4} + 36 x^{3} + 46 x^{2} + 36 x + 36)$ $4(x^2+1)(x+2)^2(9x^4+36x^3+46x^2+36x+36)$

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How do I factor this expression completely?

$4 x^{2} (x^{2} + 1) {(x + 2)}^{3} + 36 {(x^{2} + 1)}^{2} {(x + 2)}^{4}$ $4x^2(x^2+1)(x+2)^3+36(x^2+1)^2(x+2)^4$

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Explanation:

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How do I factor this expression completely?

4x2(x2+1)(x+2)3+36(x2+1)2(x+2)44x^2(x^2+1)(x+2)^3+36(x^2+1)^2(x+2)^4

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Explanation:

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$4 x^{2} (x^{2} + 1) {(x + 2)}^{3} + 36 {(x^{2} + 1)}^{2} {(x + 2)}^{4}$ $4x^2(x^2+1)(x+2)^3+36(x^2+1)^2(x+2)^4$