Solve 15m^2+m-6

$$15 m ^ { 2 } + m - 6$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 15m^{2}+am+bm-6. To find a and b, set up a system to be solved.

$$a+b=1 ab=15\left(-6\right)=-90$$

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 -90.

$$-1,90 -2,45 -3,30 -5,18 -6,15 -9,10$$

Calculate the sum for each pair.

$$-1+90=89 -2+45=43 -3+30=27 -5+18=13 -6+15=9 -9+10=1$$

The solution is the pair that gives sum 1.

$$a=-9 b=10$$

Rewrite 15m^{2}+m-6 as \left(15m^{2}-9m\right)+\left(10m-6\right).

$$\left(15m^{2}-9m\right)+\left(10m-6\right)$$

Factor out 3m in the first and 2 in the second group.

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

Factor out common term 5m-3 by using distributive property.

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

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.

$$15m^{2}+m-6=0$$

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.

$$m=\frac{-1±\sqrt{1^{2}-4\times 15\left(-6\right)}}{2\times 15}$$

Square 1.

$$m=\frac{-1±\sqrt{1-4\times 15\left(-6\right)}}{2\times 15}$$

Multiply -4 times 15.

$$m=\frac{-1±\sqrt{1-60\left(-6\right)}}{2\times 15}$$

Multiply -60 times -6.

$$m=\frac{-1±\sqrt{1+360}}{2\times 15}$$

Add 1 to 360.

$$m=\frac{-1±\sqrt{361}}{2\times 15}$$

Take the square root of 361.

$$m=\frac{-1±19}{2\times 15}$$

Multiply 2 times 15.

$$m=\frac{-1±19}{30}$$

Now solve the equation m=\frac{-1±19}{30} when ± is plus. Add -1 to 19.

$$m=\frac{18}{30}$$

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

$$m=\frac{3}{5}$$

Now solve the equation m=\frac{-1±19}{30} when ± is minus. Subtract 19 from -1.

$$m=-\frac{20}{30}$$

Reduce the fraction \frac{-20}{30} to lowest terms by extracting and canceling out 10.

$$m=-\frac{2}{3}$$

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

$$15m^{2}+m-6=15\left(m-\frac{3}{5}\right)\left(m-\left(-\frac{2}{3}\right)\right)$$

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

$$15m^{2}+m-6=15\left(m-\frac{3}{5}\right)\left(m+\frac{2}{3}\right)$$

Subtract \frac{3}{5} from m by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$15m^{2}+m-6=15\times \frac{5m-3}{5}\left(m+\frac{2}{3}\right)$$

Add \frac{2}{3} to m by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$15m^{2}+m-6=15\times \frac{5m-3}{5}\times \frac{3m+2}{3}$$

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

$$15m^{2}+m-6=15\times \frac{\left(5m-3\right)\left(3m+2\right)}{5\times 3}$$

Multiply 5 times 3.

$$15m^{2}+m-6=15\times \frac{\left(5m-3\right)\left(3m+2\right)}{15}$$

Cancel out 15, the greatest common factor in 15 and 15.

$$15m^{2}+m-6=\left(5m-3\right)\left(3m+2\right)$$

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 15

$$x ^ 2 +\frac{1}{15}x -\frac{2}{5} = 0$$

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

$$r + s = -\frac{1}{15} rs = -\frac{2}{5}$$

Two numbers r and s sum up to -\frac{1}{15} exactly when the average of the two numbers is \frac{1}{2}*-\frac{1}{15} = -\frac{1}{30}. 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>

$$r = -\frac{1}{30} - u s = -\frac{1}{30} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{2}{5}

$$(-\frac{1}{30} - u) (-\frac{1}{30} + u) = -\frac{2}{5}$$

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

$$\frac{1}{900} - u^2 = -\frac{2}{5}$$

Simplify the expression by subtracting \frac{1}{900} on both sides

$$-u^2 = -\frac{2}{5}-\frac{1}{900} = -\frac{361}{900}$$

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

$$u^2 = \frac{361}{900} u = \pm\sqrt{\frac{361}{900}} = \pm \frac{19}{30}$$

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

$$r =-\frac{1}{30} - \frac{19}{30} = -0.667 s = -\frac{1}{30} + \frac{19}{30} = 0.600$$

Evaluate

$$15m^{2}+m-6$$