# Factorize the polynomial `f(x)=x^4+5x^3-3x^2-13x+10` and find the rational zeros.

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`f(x)= x^4 + 5x^3 -3x^2 -13x +10`

We will use substitution method to find the zeros. we will try to substitute the factors of (10) = 1, -1 , 2, -2, 5, -5, 10, -10

First we will try x= 1

`f(1)= 1+5 -3 -13 + 10 = 0`

Then, x= 1 is one of the zeros of f(x), therefore, (x-1) is one of the factor.

Now we will divide f(x) by (x-1) to find the other factors.

==> `f(x)= (x-1)(x^3 +6x^2 +3x -10` )

==> Now we will substitute with x= -2 ==> f(-2)= 0

Then, (x+2) is a factor of f(x).

==> Then, we will divide (`x^3 +6x^2 +3x -10` ) by (x+2).

==> `f(x)= (x-1)(x+2)(x^2 +4x -5)`

`` Now we will factor:

**==> f(x)= (x-1)(x+2)(x+5)(x-1)**

**==>`f(x)= (x-1)^2 (x+2)(x+5)`**

**`` **

We have to find the real zeros and the factors of f(x) = x^4 + 5x^3 - 3x^2 - 13x + 10

x^4 + 5x^3 - 3x^2 - 13x + 10 = 0

=> x^4 + 5x^3 - 3x^2 - 15x + 2x + 10 = 0

=> x^3(x + 5) - 3x( x + 5) + 2(x + 5) = 0

=> (x^3 - 3x + 2)(x + 5) = 0

=> (x^3 - x^2 + x^2 - x - 2x + 2)(x + 5) = 0

=> (x^2(x - 1) + x(x - 1) - 2(x - 1))(x + 5) = 0

=> (x-1)(x^2 + x - 2)(x + 5) = 0

=> (x - 1)(x^2 + 2x - x - 2)(x + 5) = 0

=> (x - 1)(x(x + 2) - 1(x + 2))(x + 5) = 0

=> (x - 1)(x - 1)(x + 2)(x + 5) = 0

=> (x - 1)^2(x + 2)(x + 5) = 0

**The polynomial x^4 + 5x^3 - 3x^2 - 13x + 10 = (x - 1)^2(x + 2)(x + 5) and the zeros of the polynomial are x = 1, x = -2 and x = -5.**