Solve a^2+12a+32

$$a ^ { 2 } + 12 a + 32$$

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.

$$p+q=12 pq=1\times 32=32$$

Since pq is positive, p and q have the same sign. Since p+q is positive, p and q are both positive. List all such integer pairs that give product 32.

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

Calculate the sum for each pair.

$$1+32=33 2+16=18 4+8=12$$

The solution is the pair that gives sum 12.

$$p=4 q=8$$

Rewrite a^{2}+12a+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.

$$a^{2}+12a+32=0$$

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.

$$a=\frac{-12±\sqrt{12^{2}-4\times 32}}{2}$$

Square 12.

$$a=\frac{-12±\sqrt{144-4\times 32}}{2}$$

Multiply -4 times 32.

$$a=\frac{-12±\sqrt{144-128}}{2}$$

Add 144 to -128.

$$a=\frac{-12±\sqrt{16}}{2}$$

Take the square root of 16.

$$a=\frac{-12±4}{2}$$

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

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

Divide -8 by 2.

$$a=-4$$

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

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

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}.

$$a^{2}+12a+32=\left(a-\left(-4\right)\right)\left(a-\left(-8\right)\right)$$

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

$$a^{2}+12a+32=\left(a+4\right)\left(a+8\right)$$

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.

$$x ^ 2 +12x +32 = 0$$

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

$$r + s = -12 rs = 32$$

Two numbers r and s sum up to -12 exactly when the average of the two numbers is \frac{1}{2}*-12 = -6. 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>

$$r = -6 - u s = -6 + u$$

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

$$(-6 - u) (-6 + u) = 32$$

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

$$36 - u^2 = 32$$

Simplify the expression by subtracting 36 on both sides

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

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

$$u^2 = 4 u = \pm\sqrt{4} = \pm 2$$

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

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

Evaluate

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