Solve b^2-6b-16

$$b ^ { 2 } - 6 b - 16$$

Factor

$$\left(b-8\right)\left(b+2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as b^{2}+pb+qb-16. To find p and q, set up a system to be solved.

$$p+q=-6 pq=1\left(-16\right)=-16$$

Since pq is negative, p and q have the opposite signs. Since p+q is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -16.

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

Calculate the sum for each pair.

$$1-16=-15 2-8=-6 4-4=0$$

The solution is the pair that gives sum -6.

$$p=-8 q=2$$

Rewrite b^{2}-6b-16 as \left(b^{2}-8b\right)+\left(2b-16\right).

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

Factor out b in the first and 2 in the second group.

$$b\left(b-8\right)+2\left(b-8\right)$$

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

$$\left(b-8\right)\left(b+2\right)$$

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.

$$b^{2}-6b-16=0$$

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.

$$b=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-16\right)}}{2}$$

Square -6.

$$b=\frac{-\left(-6\right)±\sqrt{36-4\left(-16\right)}}{2}$$

Multiply -4 times -16.

$$b=\frac{-\left(-6\right)±\sqrt{36+64}}{2}$$

Add 36 to 64.

$$b=\frac{-\left(-6\right)±\sqrt{100}}{2}$$

Take the square root of 100.

$$b=\frac{-\left(-6\right)±10}{2}$$

The opposite of -6 is 6.

$$b=\frac{6±10}{2}$$

Now solve the equation b=\frac{6±10}{2} when ± is plus. Add 6 to 10.

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

Divide 16 by 2.

$$b=8$$

Now solve the equation b=\frac{6±10}{2} when ± is minus. Subtract 10 from 6.

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

Divide -4 by 2.

$$b=-2$$

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 -2 for x_{2}.

$$b^{2}-6b-16=\left(b-8\right)\left(b-\left(-2\right)\right)$$

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

$$b^{2}-6b-16=\left(b-8\right)\left(b+2\right)$$

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.

$$x ^ 2 -6x -16 = 0$$

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

$$r + s = 6 rs = -16$$

Two numbers r and s sum up to 6 exactly when the average of the two numbers is \frac{1}{2}*6 = 3. 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>

$$r = 3 - u s = 3 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -16

$$(3 - u) (3 + u) = -16$$

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

$$9 - u^2 = -16$$

Simplify the expression by subtracting 9 on both sides

$$-u^2 = -16-9 = -25$$

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

$$u^2 = 25 u = \pm\sqrt{25} = \pm 5$$

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

$$r =3 - 5 = -2 s = 3 + 5 = 8$$

Evaluate

$$\left(b-8\right)\left(b+2\right)$$