Solve 8x^2-44x+60

$$8 x ^ { 2 } - 44 x + 60$$

Factor

$$4\left(x-3\right)\left(2x-5\right)$$

Solution Steps

Factor out 4.

$$4\left(2x^{2}-11x+15\right)$$

Consider 2x^{2}-11x+15. Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx+15. To find a and b, set up a system to be solved.

$$a+b=-11 ab=2\times 15=30$$

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 30.

$$-1,-30 -2,-15 -3,-10 -5,-6$$

Calculate the sum for each pair.

$$-1-30=-31 -2-15=-17 -3-10=-13 -5-6=-11$$

The solution is the pair that gives sum -11.

$$a=-6 b=-5$$

Rewrite 2x^{2}-11x+15 as \left(2x^{2}-6x\right)+\left(-5x+15\right).

$$\left(2x^{2}-6x\right)+\left(-5x+15\right)$$

Factor out 2x in the first and -5 in the second group.

$$2x\left(x-3\right)-5\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(2x-5\right)$$

Rewrite the complete factored expression.

$$4\left(x-3\right)\left(2x-5\right)$$

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.

$$8x^{2}-44x+60=0$$

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.

$$x=\frac{-\left(-44\right)±\sqrt{\left(-44\right)^{2}-4\times 8\times 60}}{2\times 8}$$

Square -44.

$$x=\frac{-\left(-44\right)±\sqrt{1936-4\times 8\times 60}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-\left(-44\right)±\sqrt{1936-32\times 60}}{2\times 8}$$

Multiply -32 times 60.

$$x=\frac{-\left(-44\right)±\sqrt{1936-1920}}{2\times 8}$$

Add 1936 to -1920.

$$x=\frac{-\left(-44\right)±\sqrt{16}}{2\times 8}$$

Take the square root of 16.

$$x=\frac{-\left(-44\right)±4}{2\times 8}$$

The opposite of -44 is 44.

$$x=\frac{44±4}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{44±4}{16}$$

Now solve the equation x=\frac{44±4}{16} when ± is plus. Add 44 to 4.

$$x=\frac{48}{16}$$

Divide 48 by 16.

$$x=3$$

Now solve the equation x=\frac{44±4}{16} when ± is minus. Subtract 4 from 44.

$$x=\frac{40}{16}$$

Reduce the fraction \frac{40}{16} to lowest terms by extracting and canceling out 8.

$$x=\frac{5}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 3 for x_{1} and \frac{5}{2} for x_{2}.

$$8x^{2}-44x+60=8\left(x-3\right)\left(x-\frac{5}{2}\right)$$

Subtract \frac{5}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$8x^{2}-44x+60=8\left(x-3\right)\times \frac{2x-5}{2}$$

Cancel out 2, the greatest common factor in 8 and 2.

$$8x^{2}-44x+60=4\left(x-3\right)\left(2x-5\right)$$

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 8

$$x ^ 2 -\frac{11}{2}x +\frac{15}{2} = 0$$

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

$$r + s = \frac{11}{2} rs = \frac{15}{2}$$

Two numbers r and s sum up to \frac{11}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{11}{2} = \frac{11}{4}. 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>

$$r = \frac{11}{4} - u s = \frac{11}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{15}{2}

$$(\frac{11}{4} - u) (\frac{11}{4} + u) = \frac{15}{2}$$

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

$$\frac{121}{16} - u^2 = \frac{15}{2}$$

Simplify the expression by subtracting \frac{121}{16} on both sides

$$-u^2 = \frac{15}{2}-\frac{121}{16} = -\frac{1}{16}$$

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

$$u^2 = \frac{1}{16} u = \pm\sqrt{\frac{1}{16}} = \pm \frac{1}{4}$$

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

$$r =\frac{11}{4} - \frac{1}{4} = 2.500 s = \frac{11}{4} + \frac{1}{4} = 3$$

Evaluate

$$4\left(x-3\right)\left(2x-5\right)$$