Solve 32n^2+60n-27

$$32 n ^ { 2 } + 60 n - 27$$

Factor

$$\left(8n-3\right)\left(4n+9\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 32n^{2}+an+bn-27. To find a and b, set up a system to be solved.

$$a+b=60 ab=32\left(-27\right)=-864$$

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 -864.

$$-1,864 -2,432 -3,288 -4,216 -6,144 -8,108 -9,96 -12,72 -16,54 -18,48 -24,36 -27,32$$

Calculate the sum for each pair.

$$-1+864=863 -2+432=430 -3+288=285 -4+216=212 -6+144=138 -8+108=100 -9+96=87 -12+72=60 -16+54=38 -18+48=30 -24+36=12 -27+32=5$$

The solution is the pair that gives sum 60.

$$a=-12 b=72$$

Rewrite 32n^{2}+60n-27 as \left(32n^{2}-12n\right)+\left(72n-27\right).

$$\left(32n^{2}-12n\right)+\left(72n-27\right)$$

Factor out 4n in the first and 9 in the second group.

$$4n\left(8n-3\right)+9\left(8n-3\right)$$

Factor out common term 8n-3 by using distributive property.

$$\left(8n-3\right)\left(4n+9\right)$$

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.

$$32n^{2}+60n-27=0$$

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.

$$n=\frac{-60±\sqrt{60^{2}-4\times 32\left(-27\right)}}{2\times 32}$$

Square 60.

$$n=\frac{-60±\sqrt{3600-4\times 32\left(-27\right)}}{2\times 32}$$

Multiply -4 times 32.

$$n=\frac{-60±\sqrt{3600-128\left(-27\right)}}{2\times 32}$$

Multiply -128 times -27.

$$n=\frac{-60±\sqrt{3600+3456}}{2\times 32}$$

Add 3600 to 3456.

$$n=\frac{-60±\sqrt{7056}}{2\times 32}$$

Take the square root of 7056.

$$n=\frac{-60±84}{2\times 32}$$

Multiply 2 times 32.

$$n=\frac{-60±84}{64}$$

Now solve the equation n=\frac{-60±84}{64} when ± is plus. Add -60 to 84.

$$n=\frac{24}{64}$$

Reduce the fraction \frac{24}{64} to lowest terms by extracting and canceling out 8.

$$n=\frac{3}{8}$$

Now solve the equation n=\frac{-60±84}{64} when ± is minus. Subtract 84 from -60.

$$n=-\frac{144}{64}$$

Reduce the fraction \frac{-144}{64} to lowest terms by extracting and canceling out 16.

$$n=-\frac{9}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3}{8} for x_{1} and -\frac{9}{4} for x_{2}.

$$32n^{2}+60n-27=32\left(n-\frac{3}{8}\right)\left(n-\left(-\frac{9}{4}\right)\right)$$

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

$$32n^{2}+60n-27=32\left(n-\frac{3}{8}\right)\left(n+\frac{9}{4}\right)$$

Subtract \frac{3}{8} from n by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$32n^{2}+60n-27=32\times \frac{8n-3}{8}\left(n+\frac{9}{4}\right)$$

Add \frac{9}{4} to n by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$32n^{2}+60n-27=32\times \frac{8n-3}{8}\times \frac{4n+9}{4}$$

Multiply \frac{8n-3}{8} times \frac{4n+9}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$32n^{2}+60n-27=32\times \frac{\left(8n-3\right)\left(4n+9\right)}{8\times 4}$$

Multiply 8 times 4.

$$32n^{2}+60n-27=32\times \frac{\left(8n-3\right)\left(4n+9\right)}{32}$$

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

$$32n^{2}+60n-27=\left(8n-3\right)\left(4n+9\right)$$

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 32

$$x ^ 2 +\frac{15}{8}x -\frac{27}{32} = 0$$

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

$$r + s = -\frac{15}{8} rs = -\frac{27}{32}$$

Two numbers r and s sum up to -\frac{15}{8} exactly when the average of the two numbers is \frac{1}{2}*-\frac{15}{8} = -\frac{15}{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>

$$r = -\frac{15}{16} - u s = -\frac{15}{16} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{27}{32}

$$(-\frac{15}{16} - u) (-\frac{15}{16} + u) = -\frac{27}{32}$$

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

$$\frac{225}{256} - u^2 = -\frac{27}{32}$$

Simplify the expression by subtracting \frac{225}{256} on both sides

$$-u^2 = -\frac{27}{32}-\frac{225}{256} = -\frac{441}{256}$$

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

$$u^2 = \frac{441}{256} u = \pm\sqrt{\frac{441}{256}} = \pm \frac{21}{16}$$

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

$$r =-\frac{15}{16} - \frac{21}{16} = -2.250 s = -\frac{15}{16} + \frac{21}{16} = 0.375$$

Evaluate

$$\left(8n-3\right)\left(4n+9\right)$$