Solve 15/4-x-x^2

$$\frac{ 15 }{ 4 } -x- { x }^{ 2 }$$

Factor

$$\frac{\left(3-2x\right)\left(2x+5\right)}{4}$$

Solution Steps

Factor out \frac{1}{4}.

$$\frac{15-4x-4x^{2}}{4}$$

Consider 15-4x-4x^{2}. Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-4x^{2}-4x+15$$

Factor the expression by grouping. First, the expression needs to be rewritten as -4x^{2}+ax+bx+15. To find a and b, set up a system to be solved.

$$a+b=-4 ab=-4\times 15=-60$$

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 -60.

$$1,-60 2,-30 3,-20 4,-15 5,-12 6,-10$$

Calculate the sum for each pair.

$$1-60=-59 2-30=-28 3-20=-17 4-15=-11 5-12=-7 6-10=-4$$

The solution is the pair that gives sum -4.

$$a=6 b=-10$$

Rewrite -4x^{2}-4x+15 as \left(-4x^{2}+6x\right)+\left(-10x+15\right).

$$\left(-4x^{2}+6x\right)+\left(-10x+15\right)$$

Factor out 2x in the first and 5 in the second group.

$$2x\left(-2x+3\right)+5\left(-2x+3\right)$$

Factor out common term -2x+3 by using distributive property.

$$\left(-2x+3\right)\left(2x+5\right)$$

Rewrite the complete factored expression.

$$\frac{\left(-2x+3\right)\left(2x+5\right)}{4}$$

Evaluate

$$\frac{15}{4}-x-x^{2}$$