Solve h^2+8h-128

$$h ^ { 2 } + 8 h - 128$$

Factor

$$\left(h-8\right)\left(h+16\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as h^{2}+ah+bh-128. To find a and b, set up a system to be solved.

$$a+b=8 ab=1\left(-128\right)=-128$$

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 -128.

$$-1,128 -2,64 -4,32 -8,16$$

Calculate the sum for each pair.

$$-1+128=127 -2+64=62 -4+32=28 -8+16=8$$

The solution is the pair that gives sum 8.

$$a=-8 b=16$$

Rewrite h^{2}+8h-128 as \left(h^{2}-8h\right)+\left(16h-128\right).

$$\left(h^{2}-8h\right)+\left(16h-128\right)$$

Factor out h in the first and 16 in the second group.

$$h\left(h-8\right)+16\left(h-8\right)$$

Factor out common term h-8 by using distributive property.

$$\left(h-8\right)\left(h+16\right)$$

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.

$$h^{2}+8h-128=0$$

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.

$$h=\frac{-8±\sqrt{8^{2}-4\left(-128\right)}}{2}$$

Square 8.

$$h=\frac{-8±\sqrt{64-4\left(-128\right)}}{2}$$

Multiply -4 times -128.

$$h=\frac{-8±\sqrt{64+512}}{2}$$

Add 64 to 512.

$$h=\frac{-8±\sqrt{576}}{2}$$

Take the square root of 576.

$$h=\frac{-8±24}{2}$$

Now solve the equation h=\frac{-8±24}{2} when ± is plus. Add -8 to 24.

$$h=\frac{16}{2}$$

Divide 16 by 2.

$$h=8$$

Now solve the equation h=\frac{-8±24}{2} when ± is minus. Subtract 24 from -8.

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

Divide -32 by 2.

$$h=-16$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 8 for x_{1} and -16 for x_{2}.

$$h^{2}+8h-128=\left(h-8\right)\left(h-\left(-16\right)\right)$$

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

$$h^{2}+8h-128=\left(h-8\right)\left(h+16\right)$$

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.

$$x ^ 2 +8x -128 = 0$$

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

$$r + s = -8 rs = -128$$

Two numbers r and s sum up to -8 exactly when the average of the two numbers is \frac{1}{2}*-8 = -4. 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>

$$r = -4 - u s = -4 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -128

$$(-4 - u) (-4 + u) = -128$$

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

$$16 - u^2 = -128$$

Simplify the expression by subtracting 16 on both sides

$$-u^2 = -128-16 = -144$$

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

$$u^2 = 144 u = \pm\sqrt{144} = \pm 12$$

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

$$r =-4 - 12 = -16 s = -4 + 12 = 8$$

Evaluate

$$\left(h-8\right)\left(h+16\right)$$