Solve 15x^2-28x-32

$$15 x ^ { 2 } - 28 x - 32$$

Factor

$$\left(3x-8\right)\left(5x+4\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 15x^{2}+ax+bx-32. To find a and b, set up a system to be solved.

$$a+b=-28 ab=15\left(-32\right)=-480$$

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 -480.

$$1,-480 2,-240 3,-160 4,-120 5,-96 6,-80 8,-60 10,-48 12,-40 15,-32 16,-30 20,-24$$

Calculate the sum for each pair.

$$1-480=-479 2-240=-238 3-160=-157 4-120=-116 5-96=-91 6-80=-74 8-60=-52 10-48=-38 12-40=-28 15-32=-17 16-30=-14 20-24=-4$$

The solution is the pair that gives sum -28.

$$a=-40 b=12$$

Rewrite 15x^{2}-28x-32 as \left(15x^{2}-40x\right)+\left(12x-32\right).

$$\left(15x^{2}-40x\right)+\left(12x-32\right)$$

Factor out 5x in the first and 4 in the second group.

$$5x\left(3x-8\right)+4\left(3x-8\right)$$

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

$$\left(3x-8\right)\left(5x+4\right)$$

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.

$$15x^{2}-28x-32=0$$

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.

$$x=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\times 15\left(-32\right)}}{2\times 15}$$

Square -28.

$$x=\frac{-\left(-28\right)±\sqrt{784-4\times 15\left(-32\right)}}{2\times 15}$$

Multiply -4 times 15.

$$x=\frac{-\left(-28\right)±\sqrt{784-60\left(-32\right)}}{2\times 15}$$

Multiply -60 times -32.

$$x=\frac{-\left(-28\right)±\sqrt{784+1920}}{2\times 15}$$

Add 784 to 1920.

$$x=\frac{-\left(-28\right)±\sqrt{2704}}{2\times 15}$$

Take the square root of 2704.

$$x=\frac{-\left(-28\right)±52}{2\times 15}$$

The opposite of -28 is 28.

$$x=\frac{28±52}{2\times 15}$$

Multiply 2 times 15.

$$x=\frac{28±52}{30}$$

Now solve the equation x=\frac{28±52}{30} when ± is plus. Add 28 to 52.

$$x=\frac{80}{30}$$

Reduce the fraction \frac{80}{30} to lowest terms by extracting and canceling out 10.

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

Now solve the equation x=\frac{28±52}{30} when ± is minus. Subtract 52 from 28.

$$x=-\frac{24}{30}$$

Reduce the fraction \frac{-24}{30} to lowest terms by extracting and canceling out 6.

$$x=-\frac{4}{5}$$

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{4}{5} for x_{2}.

$$15x^{2}-28x-32=15\left(x-\frac{8}{3}\right)\left(x-\left(-\frac{4}{5}\right)\right)$$

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

$$15x^{2}-28x-32=15\left(x-\frac{8}{3}\right)\left(x+\frac{4}{5}\right)$$

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

$$15x^{2}-28x-32=15\times \frac{3x-8}{3}\left(x+\frac{4}{5}\right)$$

Add \frac{4}{5} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$15x^{2}-28x-32=15\times \frac{3x-8}{3}\times \frac{5x+4}{5}$$

Multiply \frac{3x-8}{3} times \frac{5x+4}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$15x^{2}-28x-32=15\times \frac{\left(3x-8\right)\left(5x+4\right)}{3\times 5}$$

Multiply 3 times 5.

$$15x^{2}-28x-32=15\times \frac{\left(3x-8\right)\left(5x+4\right)}{15}$$

Cancel out 15, the greatest common factor in 15 and 15.

$$15x^{2}-28x-32=\left(3x-8\right)\left(5x+4\right)$$

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 15

$$x ^ 2 -\frac{28}{15}x -\frac{32}{15} = 0$$

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

$$r + s = \frac{28}{15} rs = -\frac{32}{15}$$

Two numbers r and s sum up to \frac{28}{15} exactly when the average of the two numbers is \frac{1}{2}*\frac{28}{15} = \frac{14}{15}. 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>

$$r = \frac{14}{15} - u s = \frac{14}{15} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{32}{15}

$$(\frac{14}{15} - u) (\frac{14}{15} + u) = -\frac{32}{15}$$

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

$$\frac{196}{225} - u^2 = -\frac{32}{15}$$

Simplify the expression by subtracting \frac{196}{225} on both sides

$$-u^2 = -\frac{32}{15}-\frac{196}{225} = -\frac{676}{225}$$

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

$$u^2 = \frac{676}{225} u = \pm\sqrt{\frac{676}{225}} = \pm \frac{26}{15}$$

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

$$r =\frac{14}{15} - \frac{26}{15} = -0.800 s = \frac{14}{15} + \frac{26}{15} = 2.667$$

Evaluate

$$\left(3x-8\right)\left(5x+4\right)$$