Solve -b^3+4b-21/8):(b-3/2)quadQ(b)=

$$- b ^ { 3 } + 4 b - \frac { 21 } { 8 } ) : ( b - \frac { 3 } { 2 } ) \quad Q ( b ) =$$

Factor

$$\frac{\left(2b-3\right)\left(7-6b-4b^{2}\right)}{8}$$

Solution Steps

Factor out \frac{1}{8}.

$$\frac{-8b^{3}+32b-21}{8}$$

Consider -8b^{3}+32b-21. By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -21 and q divides the leading coefficient -8. One such root is \frac{3}{2}. Factor the polynomial by dividing it by 2b-3.

$$\left(2b-3\right)\left(-4b^{2}-6b+7\right)$$

Rewrite the complete factored expression. Polynomial -4b^{2}-6b+7 is not factored since it does not have any rational roots.

$$\frac{\left(2b-3\right)\left(-4b^{2}-6b+7\right)}{8}$$

Evaluate

$$-b^{3}+4b-\frac{21}{8}$$