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

Jan 4, 2018

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

#### Explanation:

Multiplying out we have:

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

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

120=2×3×4×5=(-5)×(-4)×(-3)×(-2)

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

Factoring is as much an art as a science.

Jan 4, 2018

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

#### Explanation:

$\left(x + 1\right) \left(x + 2\right) \left(x + 3\right) \left(x + 4\right) - 120$

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

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

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

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

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

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