Solve 8x^2-28x-60

$$8 x ^ { 2 } - 28 x - 60$$

Factor

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

Solution Steps

Factor out 4.

$$4\left(2x^{2}-7x-15\right)$$

Consider 2x^{2}-7x-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.

$$a+b=-7 ab=2\left(-15\right)=-30$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -30.

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

Calculate the sum for each pair.

$$1-30=-29 2-15=-13 3-10=-7 5-6=-1$$

The solution is the pair that gives sum -7.

$$a=-10 b=3$$

Rewrite 2x^{2}-7x-15 as \left(2x^{2}-10x\right)+\left(3x-15\right).

$$\left(2x^{2}-10x\right)+\left(3x-15\right)$$

Factor out 2x in the first and 3 in the second group.

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

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

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

Rewrite the complete factored expression.

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

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.

$$8x^{2}-28x-60=0$$

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.

$$x=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\times 8\left(-60\right)}}{2\times 8}$$

Square -28.

$$x=\frac{-\left(-28\right)±\sqrt{784-4\times 8\left(-60\right)}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-\left(-28\right)±\sqrt{784-32\left(-60\right)}}{2\times 8}$$

Multiply -32 times -60.

$$x=\frac{-\left(-28\right)±\sqrt{784+1920}}{2\times 8}$$

Add 784 to 1920.

$$x=\frac{-\left(-28\right)±\sqrt{2704}}{2\times 8}$$

Take the square root of 2704.

$$x=\frac{-\left(-28\right)±52}{2\times 8}$$

The opposite of -28 is 28.

$$x=\frac{28±52}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{28±52}{16}$$

Now solve the equation x=\frac{28±52}{16} when ± is plus. Add 28 to 52.

$$x=\frac{80}{16}$$

Divide 80 by 16.

$$x=5$$

Now solve the equation x=\frac{28±52}{16} when ± is minus. Subtract 52 from 28.

$$x=-\frac{24}{16}$$

Reduce the fraction \frac{-24}{16} to lowest terms by extracting and canceling out 8.

$$x=-\frac{3}{2}$$

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 -\frac{3}{2} for x_{2}.

$$8x^{2}-28x-60=8\left(x-5\right)\left(x-\left(-\frac{3}{2}\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$8x^{2}-28x-60=8\left(x-5\right)\left(x+\frac{3}{2}\right)$$

Add \frac{3}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$8x^{2}-28x-60=8\left(x-5\right)\times \frac{2x+3}{2}$$

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

$$8x^{2}-28x-60=4\left(x-5\right)\left(2x+3\right)$$

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

$$x ^ 2 -\frac{7}{2}x -\frac{15}{2} = 0$$

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

$$r + s = \frac{7}{2} rs = -\frac{15}{2}$$

Two numbers r and s sum up to \frac{7}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{7}{2} = \frac{7}{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>

$$r = \frac{7}{4} - u s = \frac{7}{4} + u$$

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

$$(\frac{7}{4} - u) (\frac{7}{4} + u) = -\frac{15}{2}$$

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

$$\frac{49}{16} - u^2 = -\frac{15}{2}$$

Simplify the expression by subtracting \frac{49}{16} on both sides

$$-u^2 = -\frac{15}{2}-\frac{49}{16} = -\frac{169}{16}$$

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

$$u^2 = \frac{169}{16} u = \pm\sqrt{\frac{169}{16}} = \pm \frac{13}{4}$$

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

$$r =\frac{7}{4} - \frac{13}{4} = -1.500 s = \frac{7}{4} + \frac{13}{4} = 5$$

Evaluate

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