How do I factorise (x+1)(x+2)(x+3)(x+4)-120 ?

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How do I factorise (x+1)(x+2)(x+3)(x+4)-120 ?

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Answer 1 · 2018-01-04T12:37:52

For this one I will merely give the answer as $(x - a) (x - b) (x^{2} + c x + d)$ $(x-a)(x-b)(x^2+cx+d)$ . Use what is below to work the values out.

Explanation:

Multiplying out we have:

$(x + 1) (x + 2) (x + 3) (x + 4) - 120 = x^{4} + 10 x^{3} + 35 x^{2} + 50 x - 96$ $(x+1)(x+2)(x+3)(x+4)-120=x^4+10x^3+35x^2+50x-96$

Can you find a pair of rational (actually, integer) zeroes of this polynomial? Hint:

$120 = 2 \times 3 \times 4 \times 5 = (- 5) \times (- 4) \times (- 3) \times (- 2)$ $120=2×3×4×5=(-5)×(-4)×(-3)×(-2)$

When you get these roots, they become the values of $a$ $a$ and $b$ $b$ in the above answer. Beware that one of these is negative. Divide by those two factors $x - a$ $x-a$ , $x - b$ $x-b$ and you are left with a quadratic quotient $x^{2} + c x + d$ $x^2+cx+d$ that has no more real roots (because the discriminant $c^{2} - 4 d$ $c^2-4d$ is negative).

Factoring is as much an art as a science.

Answer 2 · 2018-01-04T15:59:57

$(x^{2} + 5 x + 16) (x + 6) (x - 1)$ $(x^2+5x+16)(x+6)(x-1)$

Explanation:

$(x + 1) (x + 2) (x + 3) (x + 4) - 120$ $(x+1)(x+2)(x+3)(x+4)-120$

= $(x^{2} + 5 x + 4) \cdot (x^{2} + 5 x + 6) - 120$ $(x^2+5x+4)*(x^2+5x+6)-120$

= ${(x^{2} + 5 x + 5)}^{2} - 1 - 120$ $(x^2+5x+5)^2-1-120$

= ${(x^{2} + 5 x + 5)}^{2} - 11^{2}$ $(x^2+5x+5)^2-11^2$

= $(x^{2} + 5 x + 5 + 11) \cdot (x^{2} + 5 x + 5 - 11)$ $(x^2+5x+5+11)*(x^2+5x+5-11)$

= $(x^{2} + 5 x + 16) \cdot (x^{2} + 5 x - 6)$ $(x^2+5x+16)*(x^2+5x-6)$

= $(x^{2} + 5 x + 16) (x + 6) (x - 1)$ $(x^2+5x+16)(x+6)(x-1)$

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How do I factorise (x+1)(x+2)(x+3)(x+4)-120 ?

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