Solve P(x)=24x^5-10x^4-3x^3+x^2

$$P ( x ) = 24 x ^ { 5 } - 10 x ^ { 4 } - 3 x ^ { 3 } + x ^ { 2 }$$

Factor

$$\left(2x-1\right)\left(4x-1\right)\left(3x+1\right)x^{2}$$

Solution Steps

Factor out x^{2}.

$$x^{2}\left(24x^{3}-10x^{2}-3x+1\right)$$

Consider 24x^{3}-10x^{2}-3x+1. By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 1 and q divides the leading coefficient 24. One such root is \frac{1}{4}. Factor the polynomial by dividing it by 4x-1.

$$\left(4x-1\right)\left(6x^{2}-x-1\right)$$

Consider 6x^{2}-x-1. Factor the expression by grouping. First, the expression needs to be rewritten as 6x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=-1 ab=6\left(-1\right)=-6$$

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. List all such integer pairs that give product -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -1.

$$a=-3 b=2$$

Rewrite 6x^{2}-x-1 as \left(6x^{2}-3x\right)+\left(2x-1\right).

$$\left(6x^{2}-3x\right)+\left(2x-1\right)$$

Factor out 3x in 6x^{2}-3x.

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

Factor out common term 2x-1 by using distributive property.

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

Rewrite the complete factored expression.

$$x^{2}\left(4x-1\right)\left(2x-1\right)\left(3x+1\right)$$

Evaluate

$$\left(2x-1\right)\left(4x-1\right)\left(3x+1\right)x^{2}$$