Solve y^3-5y^2-2y+24

$$y ^ { 3 } - 5 y ^ { 2 } - 2 y + 24$$

Factor

$$\left(y-4\right)\left(y-3\right)\left(y+2\right)$$

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 24 and q divides the leading coefficient 1. One such root is 4. Factor the polynomial by dividing it by y-4.

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

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

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

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 y^{2}-y-6 as \left(y^{2}-3y\right)+\left(2y-6\right).

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

Factor out y in the first and 2 in the second group.

$$y\left(y-3\right)+2\left(y-3\right)$$

Factor out common term y-3 by using distributive property.

$$\left(y-3\right)\left(y+2\right)$$

Rewrite the complete factored expression.

$$\left(y-4\right)\left(y-3\right)\left(y+2\right)$$

Evaluate

$$\left(y-4\right)\left(y-3\right)\left(y+2\right)$$