Solve 4x^2-4x-63

$$4 x ^ { 2 } - 4 x - 63$$

Factor

$$\left(2x-9\right)\left(2x+7\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 4x^{2}+ax+bx-63. To find a and b, set up a system to be solved.

$$a+b=-4 ab=4\left(-63\right)=-252$$

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 -252.

$$1,-252 2,-126 3,-84 4,-63 6,-42 7,-36 9,-28 12,-21 14,-18$$

Calculate the sum for each pair.

$$1-252=-251 2-126=-124 3-84=-81 4-63=-59 6-42=-36 7-36=-29 9-28=-19 12-21=-9 14-18=-4$$

The solution is the pair that gives sum -4.

$$a=-18 b=14$$

Rewrite 4x^{2}-4x-63 as \left(4x^{2}-18x\right)+\left(14x-63\right).

$$\left(4x^{2}-18x\right)+\left(14x-63\right)$$

Factor out 2x in the first and 7 in the second group.

$$2x\left(2x-9\right)+7\left(2x-9\right)$$

Factor out common term 2x-9 by using distributive property.

$$\left(2x-9\right)\left(2x+7\right)$$

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.

$$4x^{2}-4x-63=0$$

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.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 4\left(-63\right)}}{2\times 4}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 4\left(-63\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-4\right)±\sqrt{16-16\left(-63\right)}}{2\times 4}$$

Multiply -16 times -63.

$$x=\frac{-\left(-4\right)±\sqrt{16+1008}}{2\times 4}$$

Add 16 to 1008.

$$x=\frac{-\left(-4\right)±\sqrt{1024}}{2\times 4}$$

Take the square root of 1024.

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

The opposite of -4 is 4.

$$x=\frac{4±32}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{4±32}{8}$$

Now solve the equation x=\frac{4±32}{8} when ± is plus. Add 4 to 32.

$$x=\frac{36}{8}$$

Reduce the fraction \frac{36}{8} to lowest terms by extracting and canceling out 4.

$$x=\frac{9}{2}$$

Now solve the equation x=\frac{4±32}{8} when ± is minus. Subtract 32 from 4.

$$x=-\frac{28}{8}$$

Reduce the fraction \frac{-28}{8} to lowest terms by extracting and canceling out 4.

$$x=-\frac{7}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{9}{2} for x_{1} and -\frac{7}{2} for x_{2}.

$$4x^{2}-4x-63=4\left(x-\frac{9}{2}\right)\left(x-\left(-\frac{7}{2}\right)\right)$$

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

$$4x^{2}-4x-63=4\left(x-\frac{9}{2}\right)\left(x+\frac{7}{2}\right)$$

Subtract \frac{9}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$4x^{2}-4x-63=4\times \frac{2x-9}{2}\left(x+\frac{7}{2}\right)$$

Add \frac{7}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$4x^{2}-4x-63=4\times \frac{2x-9}{2}\times \frac{2x+7}{2}$$

Multiply \frac{2x-9}{2} times \frac{2x+7}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$4x^{2}-4x-63=4\times \frac{\left(2x-9\right)\left(2x+7\right)}{2\times 2}$$

Multiply 2 times 2.

$$4x^{2}-4x-63=4\times \frac{\left(2x-9\right)\left(2x+7\right)}{4}$$

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

$$4x^{2}-4x-63=\left(2x-9\right)\left(2x+7\right)$$

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 4

$$x ^ 2 -1x -\frac{63}{4} = 0$$

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

$$r + s = 1 rs = -\frac{63}{4}$$

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

$$r = \frac{1}{2} - u s = \frac{1}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{63}{4}

$$(\frac{1}{2} - u) (\frac{1}{2} + u) = -\frac{63}{4}$$

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

$$\frac{1}{4} - u^2 = -\frac{63}{4}$$

Simplify the expression by subtracting \frac{1}{4} on both sides

$$-u^2 = -\frac{63}{4}-\frac{1}{4} = -16$$

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

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

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

$$r =\frac{1}{2} - 4 = -3.500 s = \frac{1}{2} + 4 = 4.500$$

Evaluate

$$\left(2x-9\right)\left(2x+7\right)$$