Solve -9p+20+p^2

$$- 9 p + 20 + p ^ { 2 }$$

Factor

$$\left(p-5\right)\left(p-4\right)$$

Solution Steps

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

$$p^{2}-9p+20$$

Factor the expression by grouping. First, the expression needs to be rewritten as p^{2}+ap+bp+20. To find a and b, set up a system to be solved.

$$a+b=-9 ab=1\times 20=20$$

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 20.

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

Calculate the sum for each pair.

$$-1-20=-21 -2-10=-12 -4-5=-9$$

The solution is the pair that gives sum -9.

$$a=-5 b=-4$$

Rewrite p^{2}-9p+20 as \left(p^{2}-5p\right)+\left(-4p+20\right).

$$\left(p^{2}-5p\right)+\left(-4p+20\right)$$

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

$$p\left(p-5\right)-4\left(p-5\right)$$

Factor out common term p-5 by using distributive property.

$$\left(p-5\right)\left(p-4\right)$$

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.

$$p^{2}-9p+20=0$$

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.

$$p=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 20}}{2}$$

Square -9.

$$p=\frac{-\left(-9\right)±\sqrt{81-4\times 20}}{2}$$

Multiply -4 times 20.

$$p=\frac{-\left(-9\right)±\sqrt{81-80}}{2}$$

Add 81 to -80.

$$p=\frac{-\left(-9\right)±\sqrt{1}}{2}$$

Take the square root of 1.

$$p=\frac{-\left(-9\right)±1}{2}$$

The opposite of -9 is 9.

$$p=\frac{9±1}{2}$$

Now solve the equation p=\frac{9±1}{2} when ± is plus. Add 9 to 1.

$$p=\frac{10}{2}$$

Divide 10 by 2.

$$p=5$$

Now solve the equation p=\frac{9±1}{2} when ± is minus. Subtract 1 from 9.

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

Divide 8 by 2.

$$p=4$$

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 4 for x_{2}.

$$p^{2}-9p+20=\left(p-5\right)\left(p-4\right)$$

Evaluate

$$\left(p-5\right)\left(p-4\right)$$