Solve 9x^2-6x-143

$$9 x ^ { 2 } - 6 x - 143$$

Factor

$$\left(3x-13\right)\left(3x+11\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 9x^{2}+ax+bx-143. To find a and b, set up a system to be solved.

$$a+b=-6 ab=9\left(-143\right)=-1287$$

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 -1287.

$$1,-1287 3,-429 9,-143 11,-117 13,-99 33,-39$$

Calculate the sum for each pair.

$$1-1287=-1286 3-429=-426 9-143=-134 11-117=-106 13-99=-86 33-39=-6$$

The solution is the pair that gives sum -6.

$$a=-39 b=33$$

Rewrite 9x^{2}-6x-143 as \left(9x^{2}-39x\right)+\left(33x-143\right).

$$\left(9x^{2}-39x\right)+\left(33x-143\right)$$

Factor out 3x in the first and 11 in the second group.

$$3x\left(3x-13\right)+11\left(3x-13\right)$$

Factor out common term 3x-13 by using distributive property.

$$\left(3x-13\right)\left(3x+11\right)$$

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.

$$9x^{2}-6x-143=0$$

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.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 9\left(-143\right)}}{2\times 9}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 9\left(-143\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-6\right)±\sqrt{36-36\left(-143\right)}}{2\times 9}$$

Multiply -36 times -143.

$$x=\frac{-\left(-6\right)±\sqrt{36+5148}}{2\times 9}$$

Add 36 to 5148.

$$x=\frac{-\left(-6\right)±\sqrt{5184}}{2\times 9}$$

Take the square root of 5184.

$$x=\frac{-\left(-6\right)±72}{2\times 9}$$

The opposite of -6 is 6.

$$x=\frac{6±72}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{6±72}{18}$$

Now solve the equation x=\frac{6±72}{18} when ± is plus. Add 6 to 72.

$$x=\frac{78}{18}$$

Reduce the fraction \frac{78}{18} to lowest terms by extracting and canceling out 6.

$$x=\frac{13}{3}$$

Now solve the equation x=\frac{6±72}{18} when ± is minus. Subtract 72 from 6.

$$x=-\frac{66}{18}$$

Reduce the fraction \frac{-66}{18} to lowest terms by extracting and canceling out 6.

$$x=-\frac{11}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{13}{3} for x_{1} and -\frac{11}{3} for x_{2}.

$$9x^{2}-6x-143=9\left(x-\frac{13}{3}\right)\left(x-\left(-\frac{11}{3}\right)\right)$$

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

$$9x^{2}-6x-143=9\left(x-\frac{13}{3}\right)\left(x+\frac{11}{3}\right)$$

Subtract \frac{13}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$9x^{2}-6x-143=9\times \frac{3x-13}{3}\left(x+\frac{11}{3}\right)$$

Add \frac{11}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$9x^{2}-6x-143=9\times \frac{3x-13}{3}\times \frac{3x+11}{3}$$

Multiply \frac{3x-13}{3} times \frac{3x+11}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$9x^{2}-6x-143=9\times \frac{\left(3x-13\right)\left(3x+11\right)}{3\times 3}$$

Multiply 3 times 3.

$$9x^{2}-6x-143=9\times \frac{\left(3x-13\right)\left(3x+11\right)}{9}$$

Cancel out 9, the greatest common factor in 9 and 9.

$$9x^{2}-6x-143=\left(3x-13\right)\left(3x+11\right)$$

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 9

$$x ^ 2 -\frac{2}{3}x -\frac{143}{9} = 0$$

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

$$r + s = \frac{2}{3} rs = -\frac{143}{9}$$

Two numbers r and s sum up to \frac{2}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{2}{3} = \frac{1}{3}. 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>

$$r = \frac{1}{3} - u s = \frac{1}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{143}{9}

$$(\frac{1}{3} - u) (\frac{1}{3} + u) = -\frac{143}{9}$$

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

$$\frac{1}{9} - u^2 = -\frac{143}{9}$$

Simplify the expression by subtracting \frac{1}{9} on both sides

$$-u^2 = -\frac{143}{9}-\frac{1}{9} = -16$$

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.

$$r =\frac{1}{3} - 4 = -3.667 s = \frac{1}{3} + 4 = 4.333$$

Evaluate

$$\left(3x-13\right)\left(3x+11\right)$$