Solve 32m^2-68m+35

$$32 m ^ { 2 } - 68 m + 35$$

Factor

$$\left(4m-5\right)\left(8m-7\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 32m^{2}+am+bm+35. To find a and b, set up a system to be solved.

$$a+b=-68 ab=32\times 35=1120$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 1120.

$$-1,-1120 -2,-560 -4,-280 -5,-224 -7,-160 -8,-140 -10,-112 -14,-80 -16,-70 -20,-56 -28,-40 -32,-35$$

Calculate the sum for each pair.

$$-1-1120=-1121 -2-560=-562 -4-280=-284 -5-224=-229 -7-160=-167 -8-140=-148 -10-112=-122 -14-80=-94 -16-70=-86 -20-56=-76 -28-40=-68 -32-35=-67$$

The solution is the pair that gives sum -68.

$$a=-40 b=-28$$

Rewrite 32m^{2}-68m+35 as \left(32m^{2}-40m\right)+\left(-28m+35\right).

$$\left(32m^{2}-40m\right)+\left(-28m+35\right)$$

Factor out 8m in the first and -7 in the second group.

$$8m\left(4m-5\right)-7\left(4m-5\right)$$

Factor out common term 4m-5 by using distributive property.

$$\left(4m-5\right)\left(8m-7\right)$$

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.

$$32m^{2}-68m+35=0$$

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.

$$m=\frac{-\left(-68\right)±\sqrt{\left(-68\right)^{2}-4\times 32\times 35}}{2\times 32}$$

Square -68.

$$m=\frac{-\left(-68\right)±\sqrt{4624-4\times 32\times 35}}{2\times 32}$$

Multiply -4 times 32.

$$m=\frac{-\left(-68\right)±\sqrt{4624-128\times 35}}{2\times 32}$$

Multiply -128 times 35.

$$m=\frac{-\left(-68\right)±\sqrt{4624-4480}}{2\times 32}$$

Add 4624 to -4480.

$$m=\frac{-\left(-68\right)±\sqrt{144}}{2\times 32}$$

Take the square root of 144.

$$m=\frac{-\left(-68\right)±12}{2\times 32}$$

The opposite of -68 is 68.

$$m=\frac{68±12}{2\times 32}$$

Multiply 2 times 32.

$$m=\frac{68±12}{64}$$

Now solve the equation m=\frac{68±12}{64} when ± is plus. Add 68 to 12.

$$m=\frac{80}{64}$$

Reduce the fraction \frac{80}{64} to lowest terms by extracting and canceling out 16.

$$m=\frac{5}{4}$$

Now solve the equation m=\frac{68±12}{64} when ± is minus. Subtract 12 from 68.

$$m=\frac{56}{64}$$

Reduce the fraction \frac{56}{64} to lowest terms by extracting and canceling out 8.

$$m=\frac{7}{8}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{4} for x_{1} and \frac{7}{8} for x_{2}.

$$32m^{2}-68m+35=32\left(m-\frac{5}{4}\right)\left(m-\frac{7}{8}\right)$$

Subtract \frac{5}{4} from m by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$32m^{2}-68m+35=32\times \frac{4m-5}{4}\left(m-\frac{7}{8}\right)$$

Subtract \frac{7}{8} from m by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$32m^{2}-68m+35=32\times \frac{4m-5}{4}\times \frac{8m-7}{8}$$

Multiply \frac{4m-5}{4} times \frac{8m-7}{8} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$32m^{2}-68m+35=32\times \frac{\left(4m-5\right)\left(8m-7\right)}{4\times 8}$$

Multiply 4 times 8.

$$32m^{2}-68m+35=32\times \frac{\left(4m-5\right)\left(8m-7\right)}{32}$$

Cancel out 32, the greatest common factor in 32 and 32.

$$32m^{2}-68m+35=\left(4m-5\right)\left(8m-7\right)$$

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 32

$$x ^ 2 -\frac{17}{8}x +\frac{35}{32} = 0$$

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

$$r + s = \frac{17}{8} rs = \frac{35}{32}$$

Two numbers r and s sum up to \frac{17}{8} exactly when the average of the two numbers is \frac{1}{2}*\frac{17}{8} = \frac{17}{16}. 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>

$$r = \frac{17}{16} - u s = \frac{17}{16} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{35}{32}

$$(\frac{17}{16} - u) (\frac{17}{16} + u) = \frac{35}{32}$$

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

$$\frac{289}{256} - u^2 = \frac{35}{32}$$

Simplify the expression by subtracting \frac{289}{256} on both sides

$$-u^2 = \frac{35}{32}-\frac{289}{256} = -\frac{9}{256}$$

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

$$u^2 = \frac{9}{256} u = \pm\sqrt{\frac{9}{256}} = \pm \frac{3}{16}$$

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

$$r =\frac{17}{16} - \frac{3}{16} = 0.875 s = \frac{17}{16} + \frac{3}{16} = 1.250$$

Evaluate

$$\left(4m-5\right)\left(8m-7\right)$$