Solve P(x)=x^3-21/4x+5/2

$$P ( x ) = x ^ { 3 } - \frac { 21 } { 4 } x + \frac { 5 } { 2 }$$

Factor

$$\frac{\left(2x-1\right)\left(2x+5\right)\left(\frac{x}{2}-1\right)}{2}$$

Solution Steps

Factor out \frac{1}{4}.

$$\frac{4x^{3}-21x+10}{4}$$

Consider 4x^{3}-21x+10. By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 10 and q divides the leading coefficient 4. One such root is -\frac{5}{2}. Factor the polynomial by dividing it by 2x+5.

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

Consider 2x^{2}-5x+2. Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx+2. To find a and b, set up a system to be solved.

$$a+b=-5 ab=2\times 2=4$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 4.

$$-1,-4 -2,-2$$

Calculate the sum for each pair.

$$-1-4=-5 -2-2=-4$$

The solution is the pair that gives sum -5.

$$a=-4 b=-1$$

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

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

Factor out 2x in the first and -1 in the second group.

$$2x\left(x-2\right)-\left(x-2\right)$$

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

$$\left(x-2\right)\left(2x-1\right)$$

Rewrite the complete factored expression.

$$\frac{\left(2x+5\right)\left(x-2\right)\left(2x-1\right)}{4}$$

Evaluate

$$x^{3}-\frac{21x}{4}+\frac{5}{2}$$