Solve 32a^2+20a-3

$$32 a ^ { 2 } + 20 a - 3$$

Factor

$$\left(8a-1\right)\left(4a+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 32a^{2}+pa+qa-3. To find p and q, set up a system to be solved.

$$p+q=20 pq=32\left(-3\right)=-96$$

Since pq is negative, p and q have the opposite signs. Since p+q is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -96.

$$-1,96 -2,48 -3,32 -4,24 -6,16 -8,12$$

Calculate the sum for each pair.

$$-1+96=95 -2+48=46 -3+32=29 -4+24=20 -6+16=10 -8+12=4$$

The solution is the pair that gives sum 20.

$$p=-4 q=24$$

Rewrite 32a^{2}+20a-3 as \left(32a^{2}-4a\right)+\left(24a-3\right).

$$\left(32a^{2}-4a\right)+\left(24a-3\right)$$

Factor out 4a in the first and 3 in the second group.

$$4a\left(8a-1\right)+3\left(8a-1\right)$$

Factor out common term 8a-1 by using distributive property.

$$\left(8a-1\right)\left(4a+3\right)$$

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.

$$32a^{2}+20a-3=0$$

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.

$$a=\frac{-20±\sqrt{20^{2}-4\times 32\left(-3\right)}}{2\times 32}$$

Square 20.

$$a=\frac{-20±\sqrt{400-4\times 32\left(-3\right)}}{2\times 32}$$

Multiply -4 times 32.

$$a=\frac{-20±\sqrt{400-128\left(-3\right)}}{2\times 32}$$

Multiply -128 times -3.

$$a=\frac{-20±\sqrt{400+384}}{2\times 32}$$

Add 400 to 384.

$$a=\frac{-20±\sqrt{784}}{2\times 32}$$

Take the square root of 784.

$$a=\frac{-20±28}{2\times 32}$$

Multiply 2 times 32.

$$a=\frac{-20±28}{64}$$

Now solve the equation a=\frac{-20±28}{64} when ± is plus. Add -20 to 28.

$$a=\frac{8}{64}$$

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

$$a=\frac{1}{8}$$

Now solve the equation a=\frac{-20±28}{64} when ± is minus. Subtract 28 from -20.

$$a=-\frac{48}{64}$$

Reduce the fraction \frac{-48}{64} to lowest terms by extracting and canceling out 16.

$$a=-\frac{3}{4}$$

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

$$32a^{2}+20a-3=32\left(a-\frac{1}{8}\right)\left(a-\left(-\frac{3}{4}\right)\right)$$

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

$$32a^{2}+20a-3=32\left(a-\frac{1}{8}\right)\left(a+\frac{3}{4}\right)$$

Subtract \frac{1}{8} from a by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$32a^{2}+20a-3=32\times \frac{8a-1}{8}\left(a+\frac{3}{4}\right)$$

Add \frac{3}{4} to a by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$32a^{2}+20a-3=32\times \frac{8a-1}{8}\times \frac{4a+3}{4}$$

Multiply \frac{8a-1}{8} times \frac{4a+3}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$32a^{2}+20a-3=32\times \frac{\left(8a-1\right)\left(4a+3\right)}{8\times 4}$$

Multiply 8 times 4.

$$32a^{2}+20a-3=32\times \frac{\left(8a-1\right)\left(4a+3\right)}{32}$$

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

$$32a^{2}+20a-3=\left(8a-1\right)\left(4a+3\right)$$

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

$$x ^ 2 +\frac{5}{8}x -\frac{3}{32} = 0$$

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

$$r + s = -\frac{5}{8} rs = -\frac{3}{32}$$

Two numbers r and s sum up to -\frac{5}{8} exactly when the average of the two numbers is \frac{1}{2}*-\frac{5}{8} = -\frac{5}{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>

$$r = -\frac{5}{16} - u s = -\frac{5}{16} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{32}

$$(-\frac{5}{16} - u) (-\frac{5}{16} + u) = -\frac{3}{32}$$

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

$$\frac{25}{256} - u^2 = -\frac{3}{32}$$

Simplify the expression by subtracting \frac{25}{256} on both sides

$$-u^2 = -\frac{3}{32}-\frac{25}{256} = -\frac{49}{256}$$

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

$$u^2 = \frac{49}{256} u = \pm\sqrt{\frac{49}{256}} = \pm \frac{7}{16}$$

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

$$r =-\frac{5}{16} - \frac{7}{16} = -0.750 s = -\frac{5}{16} + \frac{7}{16} = 0.125$$

Evaluate

$$\left(8a-1\right)\left(4a+3\right)$$