Solve x^2+40x+384

$${ x }^{ 2 } +40x+384$$

Factor

$$\left(x+16\right)\left(x+24\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+384. To find a and b, set up a system to be solved.

$$a+b=40 ab=1\times 384=384$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 384.

$$1,384 2,192 3,128 4,96 6,64 8,48 12,32 16,24$$

Calculate the sum for each pair.

$$1+384=385 2+192=194 3+128=131 4+96=100 6+64=70 8+48=56 12+32=44 16+24=40$$

The solution is the pair that gives sum 40.

$$a=16 b=24$$

Rewrite x^{2}+40x+384 as \left(x^{2}+16x\right)+\left(24x+384\right).

$$\left(x^{2}+16x\right)+\left(24x+384\right)$$

Factor out x in the first and 24 in the second group.

$$x\left(x+16\right)+24\left(x+16\right)$$

Factor out common term x+16 by using distributive property.

$$\left(x+16\right)\left(x+24\right)$$

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.

$$x^{2}+40x+384=0$$

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.

$$x=\frac{-40±\sqrt{40^{2}-4\times 384}}{2}$$

Square 40.

$$x=\frac{-40±\sqrt{1600-4\times 384}}{2}$$

Multiply -4 times 384.

$$x=\frac{-40±\sqrt{1600-1536}}{2}$$

Add 1600 to -1536.

$$x=\frac{-40±\sqrt{64}}{2}$$

Take the square root of 64.

$$x=\frac{-40±8}{2}$$

Now solve the equation x=\frac{-40±8}{2} when ± is plus. Add -40 to 8.

$$x=-\frac{32}{2}$$

Divide -32 by 2.

$$x=-16$$

Now solve the equation x=\frac{-40±8}{2} when ± is minus. Subtract 8 from -40.

$$x=-\frac{48}{2}$$

Divide -48 by 2.

$$x=-24$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -16 for x_{1} and -24 for x_{2}.

$$x^{2}+40x+384=\left(x-\left(-16\right)\right)\left(x-\left(-24\right)\right)$$

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

$$x^{2}+40x+384=\left(x+16\right)\left(x+24\right)$$

Evaluate

$$\left(x+16\right)\left(x+24\right)$$