Solve P(x)=2x^4-13x^3+7x^2+37x+15

$$P ( x ) = 2 x ^ { 4 } - 13 x ^ { 3 } + 7 x ^ { 2 } + 37 x + 15$$

Factor

$$\left(x-5\right)\left(x-3\right)\left(x+1\right)\left(2x+1\right)$$

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 15 and q divides the leading coefficient 2. One such root is -\frac{1}{2}. Factor the polynomial by dividing it by 2x+1.

$$\left(2x+1\right)\left(x^{3}-7x^{2}+7x+15\right)$$

Consider x^{3}-7x^{2}+7x+15. By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 15 and q divides the leading coefficient 1. One such root is 5. Factor the polynomial by dividing it by x-5.

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

Consider x^{2}-2x-3. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=-2 ab=1\left(-3\right)=-3$$

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

$$a=-3 b=1$$

Rewrite x^{2}-2x-3 as \left(x^{2}-3x\right)+\left(x-3\right).

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

Factor out x in x^{2}-3x.

$$x\left(x-3\right)+x-3$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(x+1\right)$$

Rewrite the complete factored expression.

$$\left(x-5\right)\left(x-3\right)\left(x+1\right)\left(2x+1\right)$$

Evaluate

$$\left(x-5\right)\left(x-3\right)\left(x+1\right)\left(2x+1\right)$$