Solve f(x)=2x^4+11x^3+15x^2-8x-20

$$f ( x ) = 2 x ^ { 4 } + 11 x ^ { 3 } + 15 x ^ { 2 } - 8 x - 20$$

Factor

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

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -20 and q divides the leading coefficient 2. One such root is -\frac{5}{2}. Factor the polynomial by dividing it by 2x+5.

$$\left(2x+5\right)\left(x^{3}+3x^{2}-4\right)$$

Consider x^{3}+3x^{2}-4. By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -4 and q divides the leading coefficient 1. One such root is -2. Factor the polynomial by dividing it by x+2.

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

Consider x^{2}+x-2. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-2. To find a and b, set up a system to be solved.

$$a+b=1 ab=1\left(-2\right)=-2$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1 b=2$$

Rewrite x^{2}+x-2 as \left(x^{2}-x\right)+\left(2x-2\right).

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

Factor out x in the first and 2 in the second group.

$$x\left(x-1\right)+2\left(x-1\right)$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(x+2\right)$$

Rewrite the complete factored expression.

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

Evaluate

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