# Mathematics

Factorize x^3+2x^2-11x-12

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1. let f(x) = x^3 + 2x^2 - 11x - 12
try x = ±1, ±2, ±3, ±4
f(1) ≠ 0
f(-1) = -1 + 2 + 11-12 = 0
So x+1 is a factor, by synthetic division:

x^3 + 2x^2 - 11x - 12
= (x+1)(x^2 + x - 12)
= (x+1)(x+4)(x-3)

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2. P ( x ) = x ^ 3 + 2 x ^ 2 - 11 x - 12

This is a polynomial of degree 3.

To find zeros for polynomials of degree 3 or higher we use Rational root test.

The Rational Root Theorem tells you that if the polynomial has a rational zero then it must be a fraction p / q, where p is a factor of the trailing constant and q is a factor of the leading coefficient.

The factor of the leading coefficient (1) is 1

The factors of the constant term ( - 12 ) are 1 , 2 , 3 , 4 , 6 and 12

Then the Rational Roots Tests yields the following possible solutions:

± 1 / 1, ± 2 / 1, ± 3 / 1, ± 4 / 1, ± 6 / 1, ± 12 / 1

If you plug these values into the polynomial P(x), you obtain P ( − 1 ) = 0

Now divide P(x) with [ x - ( - 1 ) ]

[ x - ( - 1 ) ] = ( x + 1 )

( x ^ 3 + 2 x ^ 2 - 11 x - 12 ) / ( x + 1 ) = x ^ 2 + x - 12

Polynomial x ^ 2 + x −12 can be used to find the remaining roots.

x ^ 2 + x - 12 = 0

The solutions are : x1 = - 4 and x2 = 3

Use formula for factoring quadratic equation:

a x ^ 2 + b x + c = a ( x − x1 ) ( x − x2 )

The leading coefficient is 1 so:

x ^ 2 + x - 12 = 1 [ x − ( - 4 ) ] ( x − 3 ) = ( x + 4 ) ( x - 3 )

x ^ 3 + 2 x ^ 2 - 11 x - 12 = ( x + 1 ) ( x + 4 ) ( x - 3 )

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