Solve -10m^2-39m+55

$$- 10 m ^ { 2 } - 39 m + 55$$

Factor

$$-\left(10m-11\right)\left(m+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -10m^{2}+am+bm+55. To find a and b, set up a system to be solved.

$$a+b=-39 ab=-10\times 55=-550$$

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 -550.

$$1,-550 2,-275 5,-110 10,-55 11,-50 22,-25$$

Calculate the sum for each pair.

$$1-550=-549 2-275=-273 5-110=-105 10-55=-45 11-50=-39 22-25=-3$$

The solution is the pair that gives sum -39.

$$a=11 b=-50$$

Rewrite -10m^{2}-39m+55 as \left(-10m^{2}+11m\right)+\left(-50m+55\right).

$$\left(-10m^{2}+11m\right)+\left(-50m+55\right)$$

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

$$-m\left(10m-11\right)-5\left(10m-11\right)$$

Factor out common term 10m-11 by using distributive property.

$$\left(10m-11\right)\left(-m-5\right)$$

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.

$$-10m^{2}-39m+55=0$$

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.

$$m=\frac{-\left(-39\right)±\sqrt{\left(-39\right)^{2}-4\left(-10\right)\times 55}}{2\left(-10\right)}$$

Square -39.

$$m=\frac{-\left(-39\right)±\sqrt{1521-4\left(-10\right)\times 55}}{2\left(-10\right)}$$

Multiply -4 times -10.

$$m=\frac{-\left(-39\right)±\sqrt{1521+40\times 55}}{2\left(-10\right)}$$

Multiply 40 times 55.

$$m=\frac{-\left(-39\right)±\sqrt{1521+2200}}{2\left(-10\right)}$$

Add 1521 to 2200.

$$m=\frac{-\left(-39\right)±\sqrt{3721}}{2\left(-10\right)}$$

Take the square root of 3721.

$$m=\frac{-\left(-39\right)±61}{2\left(-10\right)}$$

The opposite of -39 is 39.

$$m=\frac{39±61}{2\left(-10\right)}$$

Multiply 2 times -10.

$$m=\frac{39±61}{-20}$$

Now solve the equation m=\frac{39±61}{-20} when ± is plus. Add 39 to 61.

$$m=\frac{100}{-20}$$

Divide 100 by -20.

$$m=-5$$

Now solve the equation m=\frac{39±61}{-20} when ± is minus. Subtract 61 from 39.

$$m=-\frac{22}{-20}$$

Reduce the fraction \frac{-22}{-20} to lowest terms by extracting and canceling out 2.

$$m=\frac{11}{10}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -5 for x_{1} and \frac{11}{10} for x_{2}.

$$-10m^{2}-39m+55=-10\left(m-\left(-5\right)\right)\left(m-\frac{11}{10}\right)$$

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

$$-10m^{2}-39m+55=-10\left(m+5\right)\left(m-\frac{11}{10}\right)$$

Subtract \frac{11}{10} from m by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-10m^{2}-39m+55=-10\left(m+5\right)\times \frac{-10m+11}{-10}$$

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

$$-10m^{2}-39m+55=\left(m+5\right)\left(-10m+11\right)$$

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.

$$x ^ 2 +\frac{39}{10}x -\frac{11}{2} = 0$$

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

$$r + s = -\frac{39}{10} rs = -\frac{11}{2}$$

Two numbers r and s sum up to -\frac{39}{10} exactly when the average of the two numbers is \frac{1}{2}*-\frac{39}{10} = -\frac{39}{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>

$$r = -\frac{39}{20} - u s = -\frac{39}{20} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{11}{2}

$$(-\frac{39}{20} - u) (-\frac{39}{20} + u) = -\frac{11}{2}$$

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

$$\frac{1521}{400} - u^2 = -\frac{11}{2}$$

Simplify the expression by subtracting \frac{1521}{400} on both sides

$$-u^2 = -\frac{11}{2}-\frac{1521}{400} = -\frac{3721}{400}$$

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

$$u^2 = \frac{3721}{400} u = \pm\sqrt{\frac{3721}{400}} = \pm \frac{61}{20}$$

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

$$r =-\frac{39}{20} - \frac{61}{20} = -5 s = -\frac{39}{20} + \frac{61}{20} = 1.100$$

Evaluate

$$-\left(10m-11\right)\left(m+5\right)$$