Solve f(x)=4x^3-16x^2-23x+15

$$f ( x ) = 4 x ^ { 3 } - 16 x ^ { 2 } - 23 x + 15$$

Factor

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

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 4. One such root is -\frac{3}{2}. Factor the polynomial by dividing it by 2x+3.

$$\left(2x+3\right)\left(2x^{2}-11x+5\right)$$

Consider 2x^{2}-11x+5. Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=-11 ab=2\times 5=10$$

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 10.

$$-1,-10 -2,-5$$

Calculate the sum for each pair.

$$-1-10=-11 -2-5=-7$$

The solution is the pair that gives sum -11.

$$a=-10 b=-1$$

Rewrite 2x^{2}-11x+5 as \left(2x^{2}-10x\right)+\left(-x+5\right).

$$\left(2x^{2}-10x\right)+\left(-x+5\right)$$

Factor out 2x in the first and -1 in the second group.

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

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

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

Rewrite the complete factored expression.

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

Evaluate

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