How do you factor ${(w^{2} - 1)}^{2} - 23 (w^{2} - 1) + 120$ $(w^2-1)^2-23(w^2-1)+120$ ?

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Answer 1 · 2016-04-28T12:56:07

${(w^{2} - 1)}^{2} - 23 (w^{2} - 1) + 120 = (w + 3) (w - 3) (w + 4) (w - 4)$ $(w^2-1)^2-23(w^2-1)+120=(w+3)(w-3)(w+4)(w-4)$

${(w^{2} - 1)}^{2} - 23 (w^{2} - 1) + 120$ $(w^2-1)^2-23(w^2-1)+120$

= ${(w^{2} - 1)}^{2} - 15 (w^{2} - 1) - 8 (w^{2} - 1) + 120$ $(w^2-1)^2-15(w^2-1)-8(w^2-1)+120$

= $(w^{2} - 1) ((w^{2} - 1) - 15) - 8 ((w^{2} - 1) - 15)$ $(w^2-1)((w^2-1)-15)-8((w^2-1)-15)$

= $((w^{2} - 1) - 8) ((w^{2} - 1) - 15)$ $((w^2-1)-8)((w^2-1)-15)$

= $(w^{2} - 9) (w^{2} - 16)$ $(w^2-9)(w^2-16)$

but as $a^{2} - b^{2} = (a + b) (a - b)$ $a^2-b^2=(a+b)(a-b)$ , above is equal to

$(w + 3) (w - 3) (w + 4) (w - 4)$ $(w+3)(w-3)(w+4)(w-4)$

How do you factor (w2−1)2−23(w2−1)+120(w^2-1)^2-23(w^2-1)+120?