Solve 4x^3-40x^2+131x-140

$$4 x ^ { 3 } - 40 x ^ { 2 } + 131 x - 140$$

Factor

$$2\left(2x-7\right)\left(2x-5\right)\left(\frac{x}{2}-2\right)$$

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 -140 and q divides the leading coefficient 4. One such root is \frac{7}{2}. Factor the polynomial by dividing it by 2x-7.

$$\left(2x-7\right)\left(2x^{2}-13x+20\right)$$

Consider 2x^{2}-13x+20. Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx+20. To find a and b, set up a system to be solved.

$$a+b=-13 ab=2\times 20=40$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 40.

$$-1,-40 -2,-20 -4,-10 -5,-8$$

Calculate the sum for each pair.

$$-1-40=-41 -2-20=-22 -4-10=-14 -5-8=-13$$

The solution is the pair that gives sum -13.

$$a=-8 b=-5$$

Rewrite 2x^{2}-13x+20 as \left(2x^{2}-8x\right)+\left(-5x+20\right).

$$\left(2x^{2}-8x\right)+\left(-5x+20\right)$$

Factor out 2x in the first and -5 in the second group.

$$2x\left(x-4\right)-5\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(2x-5\right)$$

Rewrite the complete factored expression.

$$\left(2x-7\right)\left(2x-5\right)\left(x-4\right)$$

Evaluate

$$4x^{3}-40x^{2}+131x-140$$