Solve 2b^3+5b^2-4b-3

$$2 b ^ { 3 } + 5 b ^ { 2 } - 4 b - 3$$

Factor

$$\left(b-1\right)\left(b+3\right)\left(2b+1\right)$$

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 -3 and q divides the leading coefficient 2. One such root is -\frac{1}{2}. Factor the polynomial by dividing it by 2b+1.

$$\left(2b+1\right)\left(b^{2}+2b-3\right)$$

Consider b^{2}+2b-3. Factor the expression by grouping. First, the expression needs to be rewritten as b^{2}+pb+qb-3. To find p and q, set up a system to be solved.

$$p+q=2 pq=1\left(-3\right)=-3$$

Since pq is negative, p and q have the opposite signs. Since p+q is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$p=-1 q=3$$

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

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

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

$$b\left(b-1\right)+3\left(b-1\right)$$

Factor out common term b-1 by using distributive property.

$$\left(b-1\right)\left(b+3\right)$$

Rewrite the complete factored expression.

$$\left(b-1\right)\left(2b+1\right)\left(b+3\right)$$

Evaluate

$$\left(b-1\right)\left(b+3\right)\left(2b+1\right)$$