Solve 4x^2+4x-15=

$$4 { x }^{ 2 } +4x-15=$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 4x^{2}+ax+bx-15. To find a and b, set up a system to be solved.

$$a+b=4 ab=4\left(-15\right)=-60$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -60.

$$-1,60 -2,30 -3,20 -4,15 -5,12 -6,10$$

Calculate the sum for each pair.

$$-1+60=59 -2+30=28 -3+20=17 -4+15=11 -5+12=7 -6+10=4$$

The solution is the pair that gives sum 4.

$$a=-6 b=10$$

Rewrite 4x^{2}+4x-15 as \left(4x^{2}-6x\right)+\left(10x-15\right).

$$\left(4x^{2}-6x\right)+\left(10x-15\right)$$

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

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

Factor out common term 2x-3 by using distributive property.

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

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.

$$4x^{2}+4x-15=0$$

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.

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

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 4\left(-15\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-4±\sqrt{16-16\left(-15\right)}}{2\times 4}$$

Multiply -16 times -15.

$$x=\frac{-4±\sqrt{16+240}}{2\times 4}$$

Add 16 to 240.

$$x=\frac{-4±\sqrt{256}}{2\times 4}$$

Take the square root of 256.

$$x=\frac{-4±16}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-4±16}{8}$$

Now solve the equation x=\frac{-4±16}{8} when ± is plus. Add -4 to 16.

$$x=\frac{12}{8}$$

Reduce the fraction \frac{12}{8} to lowest terms by extracting and canceling out 4.

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

Now solve the equation x=\frac{-4±16}{8} when ± is minus. Subtract 16 from -4.

$$x=-\frac{20}{8}$$

Reduce the fraction \frac{-20}{8} to lowest terms by extracting and canceling out 4.

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

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

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

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

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

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

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

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

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

Multiply \frac{2x-3}{2} times \frac{2x+5}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

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

Multiply 2 times 2.

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

Cancel out 4, the greatest common factor in 4 and 4.

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

Evaluate

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