Solve 12n^2-32n-35

$$12 n ^ { 2 } - 32 n - 35$$

Factor

$$\left(2n-7\right)\left(6n+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 12n^{2}+an+bn-35. To find a and b, set up a system to be solved.

$$a+b=-32 ab=12\left(-35\right)=-420$$

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 -420.

$$1,-420 2,-210 3,-140 4,-105 5,-84 6,-70 7,-60 10,-42 12,-35 14,-30 15,-28 20,-21$$

Calculate the sum for each pair.

$$1-420=-419 2-210=-208 3-140=-137 4-105=-101 5-84=-79 6-70=-64 7-60=-53 10-42=-32 12-35=-23 14-30=-16 15-28=-13 20-21=-1$$

The solution is the pair that gives sum -32.

$$a=-42 b=10$$

Rewrite 12n^{2}-32n-35 as \left(12n^{2}-42n\right)+\left(10n-35\right).

$$\left(12n^{2}-42n\right)+\left(10n-35\right)$$

Factor out 6n in the first and 5 in the second group.

$$6n\left(2n-7\right)+5\left(2n-7\right)$$

Factor out common term 2n-7 by using distributive property.

$$\left(2n-7\right)\left(6n+5\right)$$

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.

$$12n^{2}-32n-35=0$$

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.

$$n=\frac{-\left(-32\right)±\sqrt{\left(-32\right)^{2}-4\times 12\left(-35\right)}}{2\times 12}$$

Square -32.

$$n=\frac{-\left(-32\right)±\sqrt{1024-4\times 12\left(-35\right)}}{2\times 12}$$

Multiply -4 times 12.

$$n=\frac{-\left(-32\right)±\sqrt{1024-48\left(-35\right)}}{2\times 12}$$

Multiply -48 times -35.

$$n=\frac{-\left(-32\right)±\sqrt{1024+1680}}{2\times 12}$$

Add 1024 to 1680.

$$n=\frac{-\left(-32\right)±\sqrt{2704}}{2\times 12}$$

Take the square root of 2704.

$$n=\frac{-\left(-32\right)±52}{2\times 12}$$

The opposite of -32 is 32.

$$n=\frac{32±52}{2\times 12}$$

Multiply 2 times 12.

$$n=\frac{32±52}{24}$$

Now solve the equation n=\frac{32±52}{24} when ± is plus. Add 32 to 52.

$$n=\frac{84}{24}$$

Reduce the fraction \frac{84}{24} to lowest terms by extracting and canceling out 12.

$$n=\frac{7}{2}$$

Now solve the equation n=\frac{32±52}{24} when ± is minus. Subtract 52 from 32.

$$n=-\frac{20}{24}$$

Reduce the fraction \frac{-20}{24} to lowest terms by extracting and canceling out 4.

$$n=-\frac{5}{6}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{7}{2} for x_{1} and -\frac{5}{6} for x_{2}.

$$12n^{2}-32n-35=12\left(n-\frac{7}{2}\right)\left(n-\left(-\frac{5}{6}\right)\right)$$

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

$$12n^{2}-32n-35=12\left(n-\frac{7}{2}\right)\left(n+\frac{5}{6}\right)$$

Subtract \frac{7}{2} from n by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$12n^{2}-32n-35=12\times \frac{2n-7}{2}\left(n+\frac{5}{6}\right)$$

Add \frac{5}{6} to n by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$12n^{2}-32n-35=12\times \frac{2n-7}{2}\times \frac{6n+5}{6}$$

Multiply \frac{2n-7}{2} times \frac{6n+5}{6} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$12n^{2}-32n-35=12\times \frac{\left(2n-7\right)\left(6n+5\right)}{2\times 6}$$

Multiply 2 times 6.

$$12n^{2}-32n-35=12\times \frac{\left(2n-7\right)\left(6n+5\right)}{12}$$

Cancel out 12, the greatest common factor in 12 and 12.

$$12n^{2}-32n-35=\left(2n-7\right)\left(6n+5\right)$$

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 12

$$x ^ 2 -\frac{8}{3}x -\frac{35}{12} = 0$$

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

$$r + s = \frac{8}{3} rs = -\frac{35}{12}$$

Two numbers r and s sum up to \frac{8}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{8}{3} = \frac{4}{3}. 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>

$$r = \frac{4}{3} - u s = \frac{4}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{35}{12}

$$(\frac{4}{3} - u) (\frac{4}{3} + u) = -\frac{35}{12}$$

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

$$\frac{16}{9} - u^2 = -\frac{35}{12}$$

Simplify the expression by subtracting \frac{16}{9} on both sides

$$-u^2 = -\frac{35}{12}-\frac{16}{9} = -\frac{169}{36}$$

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

$$u^2 = \frac{169}{36} u = \pm\sqrt{\frac{169}{36}} = \pm \frac{13}{6}$$

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

$$r =\frac{4}{3} - \frac{13}{6} = -0.833 s = \frac{4}{3} + \frac{13}{6} = 3.500$$

Evaluate

$$\left(2n-7\right)\left(6n+5\right)$$