Solve 10m^2+89m-9

$$10 m ^ { 2 } + 89 m - 9$$

Factor

$$\left(10m-1\right)\left(m+9\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 10m^{2}+am+bm-9. To find a and b, set up a system to be solved.

$$a+b=89 ab=10\left(-9\right)=-90$$

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 89.

$$a=-1 b=90$$

Rewrite 10m^{2}+89m-9 as \left(10m^{2}-m\right)+\left(90m-9\right).

$$\left(10m^{2}-m\right)+\left(90m-9\right)$$

Factor out m in the first and 9 in the second group.

$$m\left(10m-1\right)+9\left(10m-1\right)$$

Factor out common term 10m-1 by using distributive property.

$$\left(10m-1\right)\left(m+9\right)$$

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.

$$10m^{2}+89m-9=0$$

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.

$$m=\frac{-89±\sqrt{89^{2}-4\times 10\left(-9\right)}}{2\times 10}$$

Square 89.

$$m=\frac{-89±\sqrt{7921-4\times 10\left(-9\right)}}{2\times 10}$$

Multiply -4 times 10.

$$m=\frac{-89±\sqrt{7921-40\left(-9\right)}}{2\times 10}$$

Multiply -40 times -9.

$$m=\frac{-89±\sqrt{7921+360}}{2\times 10}$$

Add 7921 to 360.

$$m=\frac{-89±\sqrt{8281}}{2\times 10}$$

Take the square root of 8281.

$$m=\frac{-89±91}{2\times 10}$$

Multiply 2 times 10.

$$m=\frac{-89±91}{20}$$

Now solve the equation m=\frac{-89±91}{20} when ± is plus. Add -89 to 91.

$$m=\frac{2}{20}$$

Reduce the fraction \frac{2}{20} to lowest terms by extracting and canceling out 2.

$$m=\frac{1}{10}$$

Now solve the equation m=\frac{-89±91}{20} when ± is minus. Subtract 91 from -89.

$$m=-\frac{180}{20}$$

Divide -180 by 20.

$$m=-9$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1}{10} for x_{1} and -9 for x_{2}.

$$10m^{2}+89m-9=10\left(m-\frac{1}{10}\right)\left(m-\left(-9\right)\right)$$

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

$$10m^{2}+89m-9=10\left(m-\frac{1}{10}\right)\left(m+9\right)$$

Subtract \frac{1}{10} from m by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$10m^{2}+89m-9=10\times \frac{10m-1}{10}\left(m+9\right)$$

Cancel out 10, the greatest common factor in 10 and 10.

$$10m^{2}+89m-9=\left(10m-1\right)\left(m+9\right)$$

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 10

$$x ^ 2 +\frac{89}{10}x -\frac{9}{10} = 0$$

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

$$r + s = -\frac{89}{10} rs = -\frac{9}{10}$$

Two numbers r and s sum up to -\frac{89}{10} exactly when the average of the two numbers is \frac{1}{2}*-\frac{89}{10} = -\frac{89}{20}. 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>

$$r = -\frac{89}{20} - u s = -\frac{89}{20} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{9}{10}

$$(-\frac{89}{20} - u) (-\frac{89}{20} + u) = -\frac{9}{10}$$

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

$$\frac{7921}{400} - u^2 = -\frac{9}{10}$$

Simplify the expression by subtracting \frac{7921}{400} on both sides

$$-u^2 = -\frac{9}{10}-\frac{7921}{400} = -\frac{8281}{400}$$

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

$$u^2 = \frac{8281}{400} u = \pm\sqrt{\frac{8281}{400}} = \pm \frac{91}{20}$$

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

$$r =-\frac{89}{20} - \frac{91}{20} = -9 s = -\frac{89}{20} + \frac{91}{20} = 0.100$$

Evaluate

$$\left(10m-1\right)\left(m+9\right)$$