Solve p^2-10p+16

$$p ^ { 2 } - 10 p + 16$$

Factor

$$\left(p-8\right)\left(p-2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as p^{2}+ap+bp+16. To find a and b, set up a system to be solved.

$$a+b=-10 ab=1\times 16=16$$

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 16.

$$-1,-16 -2,-8 -4,-4$$

Calculate the sum for each pair.

$$-1-16=-17 -2-8=-10 -4-4=-8$$

The solution is the pair that gives sum -10.

$$a=-8 b=-2$$

Rewrite p^{2}-10p+16 as \left(p^{2}-8p\right)+\left(-2p+16\right).

$$\left(p^{2}-8p\right)+\left(-2p+16\right)$$

Factor out p in the first and -2 in the second group.

$$p\left(p-8\right)-2\left(p-8\right)$$

Factor out common term p-8 by using distributive property.

$$\left(p-8\right)\left(p-2\right)$$

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.

$$p^{2}-10p+16=0$$

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.

$$p=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 16}}{2}$$

Square -10.

$$p=\frac{-\left(-10\right)±\sqrt{100-4\times 16}}{2}$$

Multiply -4 times 16.

$$p=\frac{-\left(-10\right)±\sqrt{100-64}}{2}$$

Add 100 to -64.

$$p=\frac{-\left(-10\right)±\sqrt{36}}{2}$$

Take the square root of 36.

$$p=\frac{-\left(-10\right)±6}{2}$$

The opposite of -10 is 10.

$$p=\frac{10±6}{2}$$

Now solve the equation p=\frac{10±6}{2} when ± is plus. Add 10 to 6.

$$p=\frac{16}{2}$$

Divide 16 by 2.

$$p=8$$

Now solve the equation p=\frac{10±6}{2} when ± is minus. Subtract 6 from 10.

$$p=\frac{4}{2}$$

Divide 4 by 2.

$$p=2$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 8 for x_{1} and 2 for x_{2}.

$$p^{2}-10p+16=\left(p-8\right)\left(p-2\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.

$$x ^ 2 -10x +16 = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = 10 rs = 16$$

Two numbers r and s sum up to 10 exactly when the average of the two numbers is \frac{1}{2}*10 = 5. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = 5 - u s = 5 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 16

$$(5 - u) (5 + u) = 16$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$25 - u^2 = 16$$

Simplify the expression by subtracting 25 on both sides

$$-u^2 = 16-25 = -9$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = 9 u = \pm\sqrt{9} = \pm 3$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =5 - 3 = 2 s = 5 + 3 = 8$$

Evaluate

$$\left(p-8\right)\left(p-2\right)$$