Solve 1b^2+8b-16

$$1 b ^ { 2 } + 8 b - 16$$

Factor

$$\left(b-\left(-4\sqrt{2}-4\right)\right)\left(b-\left(4\sqrt{2}-4\right)\right)$$

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.

$$b^{2}+8b-16=0$$

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.

$$b=\frac{-8±\sqrt{8^{2}-4\left(-16\right)}}{2}$$

Square 8.

$$b=\frac{-8±\sqrt{64-4\left(-16\right)}}{2}$$

Multiply -4 times -16.

$$b=\frac{-8±\sqrt{64+64}}{2}$$

Add 64 to 64.

$$b=\frac{-8±\sqrt{128}}{2}$$

Take the square root of 128.

$$b=\frac{-8±8\sqrt{2}}{2}$$

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

$$b=\frac{8\sqrt{2}-8}{2}$$

Divide -8+8\sqrt{2} by 2.

$$b=4\sqrt{2}-4$$

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

$$b=\frac{-8\sqrt{2}-8}{2}$$

Divide -8-8\sqrt{2} by 2.

$$b=-4\sqrt{2}-4$$

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

$$b^{2}+8b-16=\left(b-\left(4\sqrt{2}-4\right)\right)\left(b-\left(-4\sqrt{2}-4\right)\right)$$

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.

$$x ^ 2 +8x -16 = 0$$

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

$$r + s = -8 rs = -16$$

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-gzdabgg4ehffg0hf.b01.azurefd.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 = -16

$$(-4 - u) (-4 + u) = -16$$

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

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

Simplify the expression by subtracting 16 on both sides

$$-u^2 = -16-16 = -32$$

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

$$u^2 = 32 u = \pm\sqrt{32} = \pm \sqrt{32}$$

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

$$r =-4 - \sqrt{32} = -9.657 s = -4 + \sqrt{32} = 1.657$$

Evaluate

$$b^{2}+8b-16$$