Solve 2y^3-5y^2-19y+42

$$2 y ^ { 3 } - 5 y ^ { 2 } - 19 y + 42$$

Factor

$$\left(y-2\right)\left(2y-7\right)\left(y+3\right)$$

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

$$\left(2y-7\right)\left(y^{2}+y-6\right)$$

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 positive, the positive number has greater absolute value than the negative. 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.

$$a=-2 b=3$$

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

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

Factor out y in the first and 3 in the second group.

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

Factor out common term y-2 by using distributive property.

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

Rewrite the complete factored expression.

$$\left(2y-7\right)\left(y-2\right)\left(y+3\right)$$

Evaluate

$$\left(y-2\right)\left(2y-7\right)\left(y+3\right)$$