Solve h(x)=2x^3+23x^2+58x-35

$$h ( x ) = 2 x ^ { 3 } + 23 x ^ { 2 } + 58 x - 35$$

Factor

$$\left(2x-1\right)\left(x+5\right)\left(x+7\right)$$

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 -35 and q divides the leading coefficient 2. One such root is \frac{1}{2}. Factor the polynomial by dividing it by 2x-1.

$$\left(2x-1\right)\left(x^{2}+12x+35\right)$$

Consider x^{2}+12x+35. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+35. To find a and b, set up a system to be solved.

$$a+b=12 ab=1\times 35=35$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 35.

$$1,35 5,7$$

Calculate the sum for each pair.

$$1+35=36 5+7=12$$

The solution is the pair that gives sum 12.

$$a=5 b=7$$

Rewrite x^{2}+12x+35 as \left(x^{2}+5x\right)+\left(7x+35\right).

$$\left(x^{2}+5x\right)+\left(7x+35\right)$$

Factor out x in the first and 7 in the second group.

$$x\left(x+5\right)+7\left(x+5\right)$$

Factor out common term x+5 by using distributive property.

$$\left(x+5\right)\left(x+7\right)$$

Rewrite the complete factored expression.

$$\left(2x-1\right)\left(x+5\right)\left(x+7\right)$$

Evaluate

$$\left(2x-1\right)\left(x+5\right)\left(x+7\right)$$