Solve p^2-84p+164

$$p ^ { 2 } - 84 p + 164$$

Factor

$$\left(p-82\right)\left(p-2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as p^{2}+ap+bp+164. To find a and b, set up a system to be solved.

$$a+b=-84 ab=1\times 164=164$$

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 164.

$$-1,-164 -2,-82 -4,-41$$

Calculate the sum for each pair.

$$-1-164=-165 -2-82=-84 -4-41=-45$$

The solution is the pair that gives sum -84.

$$a=-82 b=-2$$

Rewrite p^{2}-84p+164 as \left(p^{2}-82p\right)+\left(-2p+164\right).

$$\left(p^{2}-82p\right)+\left(-2p+164\right)$$

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

$$p\left(p-82\right)-2\left(p-82\right)$$

Factor out common term p-82 by using distributive property.

$$\left(p-82\right)\left(p-2\right)$$

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.

$$p^{2}-84p+164=0$$

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.

$$p=\frac{-\left(-84\right)±\sqrt{\left(-84\right)^{2}-4\times 164}}{2}$$

Square -84.

$$p=\frac{-\left(-84\right)±\sqrt{7056-4\times 164}}{2}$$

Multiply -4 times 164.

$$p=\frac{-\left(-84\right)±\sqrt{7056-656}}{2}$$

Add 7056 to -656.

$$p=\frac{-\left(-84\right)±\sqrt{6400}}{2}$$

Take the square root of 6400.

$$p=\frac{-\left(-84\right)±80}{2}$$

The opposite of -84 is 84.

$$p=\frac{84±80}{2}$$

Now solve the equation p=\frac{84±80}{2} when ± is plus. Add 84 to 80.

$$p=\frac{164}{2}$$

Divide 164 by 2.

$$p=82$$

Now solve the equation p=\frac{84±80}{2} when ± is minus. Subtract 80 from 84.

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

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 82 for x_{1} and 2 for x_{2}.

$$p^{2}-84p+164=\left(p-82\right)\left(p-2\right)$$

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.

$$x ^ 2 -84x +164 = 0$$

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

$$r + s = 84 rs = 164$$

Two numbers r and s sum up to 84 exactly when the average of the two numbers is \frac{1}{2}*84 = 42. 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-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = 42 - u s = 42 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 164

$$(42 - u) (42 + u) = 164$$

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

$$1764 - u^2 = 164$$

Simplify the expression by subtracting 1764 on both sides

$$-u^2 = 164-1764 = -1600$$

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

$$u^2 = 1600 u = \pm\sqrt{1600} = \pm 40$$

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

$$r =42 - 40 = 2 s = 42 + 40 = 82$$

Evaluate

$$\left(p-82\right)\left(p-2\right)$$