Question

factorise x cube minus x squared minus 17 x minus 15​

Answer 1

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Answer 2

$x^3-x^2-17x-15 = \left(x+1\right)\left(x+3\right)\left(x-5\right)$

Solution:

Given that,

We have to factorize:

$x^3 - x^2 - 17x - 15$

Substitute x = -1

$(-1)^3 - (-1)^2 - 17(-1) - 15\\\\-1 - 1 + 17 - 15\\\\-2 + 2 = 0$

Thus,  x + 1 is a factor of given polynomial

Thus the given expression becomes:

$(x+1)(x^2-2x-15)$

Now factor $x^2-2x-15$

Split the middle term

$x^2 + 3x - 5x - 15\\\\\mathrm{Break\:the\:expression\:into\:groups}\\\\\left(x^2+3x\right)+\left(-5x-15\right)\\\\\mathrm{Factor\:out\:}x\mathrm{\:from\:}x^2+3x\\\\x\left(x+3\right) + (-5x -15)\\\\\mathrm{Factor\:out\:}-5\mathrm{\:from\:}-5x-15\\\\x\left(x+3\right) -5(x + 3)\\\\\mathrm{Factor\:out\:common\:term\:}x+3\\\\\left(x+3\right)\left(x-5\right)$

Thus two other factors are (x + 3) and (x - 5)

Thus the factors of given expression are (x + 1) , (x + 3) and (x - 5)

$x^3-x^2-17x-15 = \left(x+1\right)\left(x+3\right)\left(x-5\right)$

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