Solve a^2+4a-32

$$a ^ { 2 } + 4 a - 32$$

Factor

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

Solution Steps

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

$$p+q=4 pq=1\left(-32\right)=-32$$

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 -32.

$$-1,32 -2,16 -4,8$$

Calculate the sum for each pair.

$$-1+32=31 -2+16=14 -4+8=4$$

The solution is the pair that gives sum 4.

$$p=-4 q=8$$

Rewrite a^{2}+4a-32 as \left(a^{2}-4a\right)+\left(8a-32\right).

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

Factor out a in the first and 8 in the second group.

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

Factor out common term a-4 by using distributive property.

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

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.

$$a^{2}+4a-32=0$$

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.

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

Square 4.

$$a=\frac{-4±\sqrt{16-4\left(-32\right)}}{2}$$

Multiply -4 times -32.

$$a=\frac{-4±\sqrt{16+128}}{2}$$

Add 16 to 128.

$$a=\frac{-4±\sqrt{144}}{2}$$

Take the square root of 144.

$$a=\frac{-4±12}{2}$$

Now solve the equation a=\frac{-4±12}{2} when ± is plus. Add -4 to 12.

$$a=\frac{8}{2}$$

Divide 8 by 2.

$$a=4$$

Now solve the equation a=\frac{-4±12}{2} when ± is minus. Subtract 12 from -4.

$$a=-\frac{16}{2}$$

Divide -16 by 2.

$$a=-8$$

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

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

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

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

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.

$$x ^ 2 +4x -32 = 0$$

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

$$r + s = -4 rs = -32$$

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

$$r = -2 - u s = -2 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -32

$$(-2 - u) (-2 + u) = -32$$

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

$$4 - u^2 = -32$$

Simplify the expression by subtracting 4 on both sides

$$-u^2 = -32-4 = -36$$

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

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

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

$$r =-2 - 6 = -8 s = -2 + 6 = 4$$

Evaluate

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