Solve 10m^2-21m-10

$$10 m ^ { 2 } - 21 m - 10$$

Factor

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

Solution Steps

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

$$a+b=-21 ab=10\left(-10\right)=-100$$

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 -100.

$$1,-100 2,-50 4,-25 5,-20 10,-10$$

Calculate the sum for each pair.

$$1-100=-99 2-50=-48 4-25=-21 5-20=-15 10-10=0$$

The solution is the pair that gives sum -21.

$$a=-25 b=4$$

Rewrite 10m^{2}-21m-10 as \left(10m^{2}-25m\right)+\left(4m-10\right).

$$\left(10m^{2}-25m\right)+\left(4m-10\right)$$

Factor out 5m in the first and 2 in the second group.

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

Factor out common term 2m-5 by using distributive property.

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

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.

$$10m^{2}-21m-10=0$$

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.

$$m=\frac{-\left(-21\right)±\sqrt{\left(-21\right)^{2}-4\times 10\left(-10\right)}}{2\times 10}$$

Square -21.

$$m=\frac{-\left(-21\right)±\sqrt{441-4\times 10\left(-10\right)}}{2\times 10}$$

Multiply -4 times 10.

$$m=\frac{-\left(-21\right)±\sqrt{441-40\left(-10\right)}}{2\times 10}$$

Multiply -40 times -10.

$$m=\frac{-\left(-21\right)±\sqrt{441+400}}{2\times 10}$$

Add 441 to 400.

$$m=\frac{-\left(-21\right)±\sqrt{841}}{2\times 10}$$

Take the square root of 841.

$$m=\frac{-\left(-21\right)±29}{2\times 10}$$

The opposite of -21 is 21.

$$m=\frac{21±29}{2\times 10}$$

Multiply 2 times 10.

$$m=\frac{21±29}{20}$$

Now solve the equation m=\frac{21±29}{20} when ± is plus. Add 21 to 29.

$$m=\frac{50}{20}$$

Reduce the fraction \frac{50}{20} to lowest terms by extracting and canceling out 10.

$$m=\frac{5}{2}$$

Now solve the equation m=\frac{21±29}{20} when ± is minus. Subtract 29 from 21.

$$m=-\frac{8}{20}$$

Reduce the fraction \frac{-8}{20} to lowest terms by extracting and canceling out 4.

$$m=-\frac{2}{5}$$

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

$$10m^{2}-21m-10=10\left(m-\frac{5}{2}\right)\left(m-\left(-\frac{2}{5}\right)\right)$$

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

$$10m^{2}-21m-10=10\left(m-\frac{5}{2}\right)\left(m+\frac{2}{5}\right)$$

Subtract \frac{5}{2} from m by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$10m^{2}-21m-10=10\times \frac{2m-5}{2}\left(m+\frac{2}{5}\right)$$

Add \frac{2}{5} to m by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$10m^{2}-21m-10=10\times \frac{2m-5}{2}\times \frac{5m+2}{5}$$

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

$$10m^{2}-21m-10=10\times \frac{\left(2m-5\right)\left(5m+2\right)}{2\times 5}$$

Multiply 2 times 5.

$$10m^{2}-21m-10=10\times \frac{\left(2m-5\right)\left(5m+2\right)}{10}$$

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

$$10m^{2}-21m-10=\left(2m-5\right)\left(5m+2\right)$$

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 10

$$x ^ 2 -\frac{21}{10}x -1 = 0$$

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

$$r + s = \frac{21}{10} rs = -1$$

Two numbers r and s sum up to \frac{21}{10} exactly when the average of the two numbers is \frac{1}{2}*\frac{21}{10} = \frac{21}{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>

$$r = \frac{21}{20} - u s = \frac{21}{20} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -1

$$(\frac{21}{20} - u) (\frac{21}{20} + u) = -1$$

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

$$\frac{441}{400} - u^2 = -1$$

Simplify the expression by subtracting \frac{441}{400} on both sides

$$-u^2 = -1-\frac{441}{400} = -\frac{841}{400}$$

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

$$u^2 = \frac{841}{400} u = \pm\sqrt{\frac{841}{400}} = \pm \frac{29}{20}$$

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

$$r =\frac{21}{20} - \frac{29}{20} = -0.400 s = \frac{21}{20} + \frac{29}{20} = 2.500$$

Evaluate

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