Solve 9p^2-30p+16

$$9 p ^ { 2 } - 30 p + 16$$

Factor

$$\left(3p-8\right)\left(3p-2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 9p^{2}+ap+bp+16. To find a and b, set up a system to be solved.

$$a+b=-30 ab=9\times 16=144$$

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 144.

$$-1,-144 -2,-72 -3,-48 -4,-36 -6,-24 -8,-18 -9,-16 -12,-12$$

Calculate the sum for each pair.

$$-1-144=-145 -2-72=-74 -3-48=-51 -4-36=-40 -6-24=-30 -8-18=-26 -9-16=-25 -12-12=-24$$

The solution is the pair that gives sum -30.

$$a=-24 b=-6$$

Rewrite 9p^{2}-30p+16 as \left(9p^{2}-24p\right)+\left(-6p+16\right).

$$\left(9p^{2}-24p\right)+\left(-6p+16\right)$$

Factor out 3p in the first and -2 in the second group.

$$3p\left(3p-8\right)-2\left(3p-8\right)$$

Factor out common term 3p-8 by using distributive property.

$$\left(3p-8\right)\left(3p-2\right)$$

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.

$$9p^{2}-30p+16=0$$

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.

$$p=\frac{-\left(-30\right)±\sqrt{\left(-30\right)^{2}-4\times 9\times 16}}{2\times 9}$$

Square -30.

$$p=\frac{-\left(-30\right)±\sqrt{900-4\times 9\times 16}}{2\times 9}$$

Multiply -4 times 9.

$$p=\frac{-\left(-30\right)±\sqrt{900-36\times 16}}{2\times 9}$$

Multiply -36 times 16.

$$p=\frac{-\left(-30\right)±\sqrt{900-576}}{2\times 9}$$

Add 900 to -576.

$$p=\frac{-\left(-30\right)±\sqrt{324}}{2\times 9}$$

Take the square root of 324.

$$p=\frac{-\left(-30\right)±18}{2\times 9}$$

The opposite of -30 is 30.

$$p=\frac{30±18}{2\times 9}$$

Multiply 2 times 9.

$$p=\frac{30±18}{18}$$

Now solve the equation p=\frac{30±18}{18} when ± is plus. Add 30 to 18.

$$p=\frac{48}{18}$$

Reduce the fraction \frac{48}{18} to lowest terms by extracting and canceling out 6.

$$p=\frac{8}{3}$$

Now solve the equation p=\frac{30±18}{18} when ± is minus. Subtract 18 from 30.

$$p=\frac{12}{18}$$

Reduce the fraction \frac{12}{18} to lowest terms by extracting and canceling out 6.

$$p=\frac{2}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{8}{3} for x_{1} and \frac{2}{3} for x_{2}.

$$9p^{2}-30p+16=9\left(p-\frac{8}{3}\right)\left(p-\frac{2}{3}\right)$$

Subtract \frac{8}{3} from p by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$9p^{2}-30p+16=9\times \frac{3p-8}{3}\left(p-\frac{2}{3}\right)$$

Subtract \frac{2}{3} from p by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$9p^{2}-30p+16=9\times \frac{3p-8}{3}\times \frac{3p-2}{3}$$

Multiply \frac{3p-8}{3} times \frac{3p-2}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$9p^{2}-30p+16=9\times \frac{\left(3p-8\right)\left(3p-2\right)}{3\times 3}$$

Multiply 3 times 3.

$$9p^{2}-30p+16=9\times \frac{\left(3p-8\right)\left(3p-2\right)}{9}$$

Cancel out 9, the greatest common factor in 9 and 9.

$$9p^{2}-30p+16=\left(3p-8\right)\left(3p-2\right)$$

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.This is achieved by dividing both sides of the equation by 9

$$x ^ 2 -\frac{10}{3}x +\frac{16}{9} = 0$$

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

$$r + s = \frac{10}{3} rs = \frac{16}{9}$$

Two numbers r and s sum up to \frac{10}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{10}{3} = \frac{5}{3}. 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>

$$r = \frac{5}{3} - u s = \frac{5}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{16}{9}

$$(\frac{5}{3} - u) (\frac{5}{3} + u) = \frac{16}{9}$$

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

$$\frac{25}{9} - u^2 = \frac{16}{9}$$

Simplify the expression by subtracting \frac{25}{9} on both sides

$$-u^2 = \frac{16}{9}-\frac{25}{9} = -1$$

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

$$u^2 = 1 u = \pm\sqrt{1} = \pm 1$$

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

$$r =\frac{5}{3} - 1 = 0.667 s = \frac{5}{3} + 1 = 2.667$$

Evaluate

$$\left(3p-8\right)\left(3p-2\right)$$