Solve 9m^2-9m-28

$$9 m ^ { 2 } - 9 m - 28$$

Factor

$$\left(3m-7\right)\left(3m+4\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 9m^{2}+am+bm-28. To find a and b, set up a system to be solved.

$$a+b=-9 ab=9\left(-28\right)=-252$$

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 -9.

$$a=-21 b=12$$

Rewrite 9m^{2}-9m-28 as \left(9m^{2}-21m\right)+\left(12m-28\right).

$$\left(9m^{2}-21m\right)+\left(12m-28\right)$$

Factor out 3m in the first and 4 in the second group.

$$3m\left(3m-7\right)+4\left(3m-7\right)$$

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

$$\left(3m-7\right)\left(3m+4\right)$$

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.

$$9m^{2}-9m-28=0$$

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.

$$m=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 9\left(-28\right)}}{2\times 9}$$

Square -9.

$$m=\frac{-\left(-9\right)±\sqrt{81-4\times 9\left(-28\right)}}{2\times 9}$$

Multiply -4 times 9.

$$m=\frac{-\left(-9\right)±\sqrt{81-36\left(-28\right)}}{2\times 9}$$

Multiply -36 times -28.

$$m=\frac{-\left(-9\right)±\sqrt{81+1008}}{2\times 9}$$

Add 81 to 1008.

$$m=\frac{-\left(-9\right)±\sqrt{1089}}{2\times 9}$$

Take the square root of 1089.

$$m=\frac{-\left(-9\right)±33}{2\times 9}$$

The opposite of -9 is 9.

$$m=\frac{9±33}{2\times 9}$$

Multiply 2 times 9.

$$m=\frac{9±33}{18}$$

Now solve the equation m=\frac{9±33}{18} when ± is plus. Add 9 to 33.

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

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

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

Now solve the equation m=\frac{9±33}{18} when ± is minus. Subtract 33 from 9.

$$m=-\frac{24}{18}$$

Reduce the fraction \frac{-24}{18} to lowest terms by extracting and canceling out 6.

$$m=-\frac{4}{3}$$

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{4}{3} for x_{2}.

$$9m^{2}-9m-28=9\left(m-\frac{7}{3}\right)\left(m-\left(-\frac{4}{3}\right)\right)$$

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

$$9m^{2}-9m-28=9\left(m-\frac{7}{3}\right)\left(m+\frac{4}{3}\right)$$

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

$$9m^{2}-9m-28=9\times \frac{3m-7}{3}\left(m+\frac{4}{3}\right)$$

Add \frac{4}{3} to m by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$9m^{2}-9m-28=9\times \frac{3m-7}{3}\times \frac{3m+4}{3}$$

Multiply \frac{3m-7}{3} times \frac{3m+4}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$9m^{2}-9m-28=9\times \frac{\left(3m-7\right)\left(3m+4\right)}{3\times 3}$$

Multiply 3 times 3.

$$9m^{2}-9m-28=9\times \frac{\left(3m-7\right)\left(3m+4\right)}{9}$$

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

$$9m^{2}-9m-28=\left(3m-7\right)\left(3m+4\right)$$

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 9

$$x ^ 2 -1x -\frac{28}{9} = 0$$

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

$$r + s = 1 rs = -\frac{28}{9}$$

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{28}{9}

$$(\frac{1}{2} - u) (\frac{1}{2} + u) = -\frac{28}{9}$$

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

$$\frac{1}{4} - u^2 = -\frac{28}{9}$$

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

$$-u^2 = -\frac{28}{9}-\frac{1}{4} = -\frac{121}{36}$$

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

$$u^2 = \frac{121}{36} u = \pm\sqrt{\frac{121}{36}} = \pm \frac{11}{6}$$

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

$$r =\frac{1}{2} - \frac{11}{6} = -1.333 s = \frac{1}{2} + \frac{11}{6} = 2.333$$

Evaluate

$$\left(3m-7\right)\left(3m+4\right)$$