Solve -2p^2+110+12p

$$- 2 p ^ { 2 } + 110 + 12 p$$

Factor

$$-2\left(p-11\right)\left(p+5\right)$$

Solution Steps

Factor out 2.

$$2\left(-p^{2}+55+6p\right)$$

Consider -p^{2}+55+6p. Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-p^{2}+6p+55$$

Factor the expression by grouping. First, the expression needs to be rewritten as -p^{2}+ap+bp+55. To find a and b, set up a system to be solved.

$$a+b=6 ab=-55=-55$$

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 -55.

$$-1,55 -5,11$$

Calculate the sum for each pair.

$$-1+55=54 -5+11=6$$

The solution is the pair that gives sum 6.

$$a=11 b=-5$$

Rewrite -p^{2}+6p+55 as \left(-p^{2}+11p\right)+\left(-5p+55\right).

$$\left(-p^{2}+11p\right)+\left(-5p+55\right)$$

Factor out -p in the first and -5 in the second group.

$$-p\left(p-11\right)-5\left(p-11\right)$$

Factor out common term p-11 by using distributive property.

$$\left(p-11\right)\left(-p-5\right)$$

Rewrite the complete factored expression.

$$2\left(p-11\right)\left(-p-5\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$-2p^{2}+12p+110=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$p=\frac{-12±\sqrt{12^{2}-4\left(-2\right)\times 110}}{2\left(-2\right)}$$

Square 12.

$$p=\frac{-12±\sqrt{144-4\left(-2\right)\times 110}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$p=\frac{-12±\sqrt{144+8\times 110}}{2\left(-2\right)}$$

Multiply 8 times 110.

$$p=\frac{-12±\sqrt{144+880}}{2\left(-2\right)}$$

Add 144 to 880.

$$p=\frac{-12±\sqrt{1024}}{2\left(-2\right)}$$

Take the square root of 1024.

$$p=\frac{-12±32}{2\left(-2\right)}$$

Multiply 2 times -2.

$$p=\frac{-12±32}{-4}$$

Now solve the equation p=\frac{-12±32}{-4} when ± is plus. Add -12 to 32.

$$p=\frac{20}{-4}$$

Divide 20 by -4.

$$p=-5$$

Now solve the equation p=\frac{-12±32}{-4} when ± is minus. Subtract 32 from -12.

$$p=-\frac{44}{-4}$$

Divide -44 by -4.

$$p=11$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -5 for x_{1} and 11 for x_{2}.

$$-2p^{2}+12p+110=-2\left(p-\left(-5\right)\right)\left(p-11\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$-2p^{2}+12p+110=-2\left(p+5\right)\left(p-11\right)$$

Evaluate

$$-2\left(p-11\right)\left(p+5\right)$$