Solve f(x)=12x^3+53x^2-5x-100

$$f ( x ) = 12 x ^ { 3 } + 53 x ^ { 2 } - 5 x - 100$$

Factor

$$4\left(4x-5\right)\left(3x+5\right)\left(\frac{x}{4}+1\right)$$

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 -100 and q divides the leading coefficient 12. One such root is -\frac{5}{3}. Factor the polynomial by dividing it by 3x+5.

$$\left(3x+5\right)\left(4x^{2}+11x-20\right)$$

Consider 4x^{2}+11x-20. Factor the expression by grouping. First, the expression needs to be rewritten as 4x^{2}+ax+bx-20. To find a and b, set up a system to be solved.

$$a+b=11 ab=4\left(-20\right)=-80$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -80.

$$-1,80 -2,40 -4,20 -5,16 -8,10$$

Calculate the sum for each pair.

$$-1+80=79 -2+40=38 -4+20=16 -5+16=11 -8+10=2$$

The solution is the pair that gives sum 11.

$$a=-5 b=16$$

Rewrite 4x^{2}+11x-20 as \left(4x^{2}-5x\right)+\left(16x-20\right).

$$\left(4x^{2}-5x\right)+\left(16x-20\right)$$

Factor out x in the first and 4 in the second group.

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

Factor out common term 4x-5 by using distributive property.

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

Rewrite the complete factored expression.

$$\left(4x-5\right)\left(x+4\right)\left(3x+5\right)$$

Evaluate

$$12x^{3}+53x^{2}-5x-100$$