Solve 6p+40-p^2

$$6 p + 40 - p ^ { 2 }$$

Factor

$$-\left(p-10\right)\left(p+4\right)$$

Solution Steps

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-p^{2}+6p+40$$

Factor the expression by grouping. First, the expression needs to be rewritten as -p^{2}+ap+bp+40. To find a and b, set up a system to be solved.

$$a+b=6 ab=-40=-40$$

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 -40.

$$-1,40 -2,20 -4,10 -5,8$$

Calculate the sum for each pair.

$$-1+40=39 -2+20=18 -4+10=6 -5+8=3$$

The solution is the pair that gives sum 6.

$$a=10 b=-4$$

Rewrite -p^{2}+6p+40 as \left(-p^{2}+10p\right)+\left(-4p+40\right).

$$\left(-p^{2}+10p\right)+\left(-4p+40\right)$$

Factor out -p in the first and -4 in the second group.

$$-p\left(p-10\right)-4\left(p-10\right)$$

Factor out common term p-10 by using distributive property.

$$\left(p-10\right)\left(-p-4\right)$$

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.

$$-p^{2}+6p+40=0$$

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.

$$p=\frac{-6±\sqrt{6^{2}-4\left(-1\right)\times 40}}{2\left(-1\right)}$$

Square 6.

$$p=\frac{-6±\sqrt{36-4\left(-1\right)\times 40}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$p=\frac{-6±\sqrt{36+4\times 40}}{2\left(-1\right)}$$

Multiply 4 times 40.

$$p=\frac{-6±\sqrt{36+160}}{2\left(-1\right)}$$

Add 36 to 160.

$$p=\frac{-6±\sqrt{196}}{2\left(-1\right)}$$

Take the square root of 196.

$$p=\frac{-6±14}{2\left(-1\right)}$$

Multiply 2 times -1.

$$p=\frac{-6±14}{-2}$$

Now solve the equation p=\frac{-6±14}{-2} when ± is plus. Add -6 to 14.

$$p=\frac{8}{-2}$$

Divide 8 by -2.

$$p=-4$$

Now solve the equation p=\frac{-6±14}{-2} when ± is minus. Subtract 14 from -6.

$$p=-\frac{20}{-2}$$

Divide -20 by -2.

$$p=10$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -4 for x_{1} and 10 for x_{2}.

$$-p^{2}+6p+40=-\left(p-\left(-4\right)\right)\left(p-10\right)$$

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

$$-p^{2}+6p+40=-\left(p+4\right)\left(p-10\right)$$

Evaluate

$$-\left(p-10\right)\left(p+4\right)$$