Solve 9x^2-6x-35

$$9 x ^ { 2 } - 6 x - 35$$

Factor

$$\left(3x-7\right)\left(3x+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 9x^{2}+ax+bx-35. To find a and b, set up a system to be solved.

$$a+b=-6 ab=9\left(-35\right)=-315$$

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 -315.

$$1,-315 3,-105 5,-63 7,-45 9,-35 15,-21$$

Calculate the sum for each pair.

$$1-315=-314 3-105=-102 5-63=-58 7-45=-38 9-35=-26 15-21=-6$$

The solution is the pair that gives sum -6.

$$a=-21 b=15$$

Rewrite 9x^{2}-6x-35 as \left(9x^{2}-21x\right)+\left(15x-35\right).

$$\left(9x^{2}-21x\right)+\left(15x-35\right)$$

Factor out 3x in the first and 5 in the second group.

$$3x\left(3x-7\right)+5\left(3x-7\right)$$

Factor out common term 3x-7 by using distributive property.

$$\left(3x-7\right)\left(3x+5\right)$$

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.

$$9x^{2}-6x-35=0$$

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.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 9\left(-35\right)}}{2\times 9}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 9\left(-35\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-6\right)±\sqrt{36-36\left(-35\right)}}{2\times 9}$$

Multiply -36 times -35.

$$x=\frac{-\left(-6\right)±\sqrt{36+1260}}{2\times 9}$$

Add 36 to 1260.

$$x=\frac{-\left(-6\right)±\sqrt{1296}}{2\times 9}$$

Take the square root of 1296.

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

The opposite of -6 is 6.

$$x=\frac{6±36}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{6±36}{18}$$

Now solve the equation x=\frac{6±36}{18} when ± is plus. Add 6 to 36.

$$x=\frac{42}{18}$$

Reduce the fraction \frac{42}{18} to lowest terms by extracting and canceling out 6.

$$x=\frac{7}{3}$$

Now solve the equation x=\frac{6±36}{18} when ± is minus. Subtract 36 from 6.

$$x=-\frac{30}{18}$$

Reduce the fraction \frac{-30}{18} to lowest terms by extracting and canceling out 6.

$$x=-\frac{5}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{7}{3} for x_{1} and -\frac{5}{3} for x_{2}.

$$9x^{2}-6x-35=9\left(x-\frac{7}{3}\right)\left(x-\left(-\frac{5}{3}\right)\right)$$

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

$$9x^{2}-6x-35=9\left(x-\frac{7}{3}\right)\left(x+\frac{5}{3}\right)$$

Subtract \frac{7}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$9x^{2}-6x-35=9\times \frac{3x-7}{3}\left(x+\frac{5}{3}\right)$$

Add \frac{5}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$9x^{2}-6x-35=9\times \frac{3x-7}{3}\times \frac{3x+5}{3}$$

Multiply \frac{3x-7}{3} times \frac{3x+5}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$9x^{2}-6x-35=9\times \frac{\left(3x-7\right)\left(3x+5\right)}{3\times 3}$$

Multiply 3 times 3.

$$9x^{2}-6x-35=9\times \frac{\left(3x-7\right)\left(3x+5\right)}{9}$$

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

$$9x^{2}-6x-35=\left(3x-7\right)\left(3x+5\right)$$

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

$$x ^ 2 -\frac{2}{3}x -\frac{35}{9} = 0$$

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

$$r + s = \frac{2}{3} rs = -\frac{35}{9}$$

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{35}{9}

$$(\frac{1}{3} - u) (\frac{1}{3} + u) = -\frac{35}{9}$$

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

$$\frac{1}{9} - u^2 = -\frac{35}{9}$$

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

$$-u^2 = -\frac{35}{9}-\frac{1}{9} = -4$$

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

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

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

$$r =\frac{1}{3} - 2 = -1.667 s = \frac{1}{3} + 2 = 2.333$$

Evaluate

$$\left(3x-7\right)\left(3x+5\right)$$