Solve -v^2-10v-16

$$- v ^ { 2 } - 10 v - 16$$

Factor

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

Solution Steps

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

$$a+b=-10 ab=-\left(-16\right)=16$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 16.

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

Calculate the sum for each pair.

$$-1-16=-17 -2-8=-10 -4-4=-8$$

The solution is the pair that gives sum -10.

$$a=-2 b=-8$$

Rewrite -v^{2}-10v-16 as \left(-v^{2}-2v\right)+\left(-8v-16\right).

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

Factor out v in the first and 8 in the second group.

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

Factor out common term -v-2 by using distributive property.

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

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.

$$-v^{2}-10v-16=0$$

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.

$$v=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-1\right)\left(-16\right)}}{2\left(-1\right)}$$

Square -10.

$$v=\frac{-\left(-10\right)±\sqrt{100-4\left(-1\right)\left(-16\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$v=\frac{-\left(-10\right)±\sqrt{100+4\left(-16\right)}}{2\left(-1\right)}$$

Multiply 4 times -16.

$$v=\frac{-\left(-10\right)±\sqrt{100-64}}{2\left(-1\right)}$$

Add 100 to -64.

$$v=\frac{-\left(-10\right)±\sqrt{36}}{2\left(-1\right)}$$

Take the square root of 36.

$$v=\frac{-\left(-10\right)±6}{2\left(-1\right)}$$

The opposite of -10 is 10.

$$v=\frac{10±6}{2\left(-1\right)}$$

Multiply 2 times -1.

$$v=\frac{10±6}{-2}$$

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

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

Divide 16 by -2.

$$v=-8$$

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

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

Divide 4 by -2.

$$v=-2$$

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}.

$$-v^{2}-10v-16=-\left(v-\left(-8\right)\right)\left(v-\left(-2\right)\right)$$

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

$$-v^{2}-10v-16=-\left(v+8\right)\left(v+2\right)$$

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.

$$x ^ 2 +10x +16 = 0$$

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

$$r + s = -10 rs = 16$$

Two numbers r and s sum up to -10 exactly when the average of the two numbers is \frac{1}{2}*-10 = -5. 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>

$$r = -5 - u s = -5 + u$$

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

$$(-5 - u) (-5 + u) = 16$$

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

$$25 - u^2 = 16$$

Simplify the expression by subtracting 25 on both sides

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

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

$$u^2 = 9 u = \pm\sqrt{9} = \pm 3$$

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

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

Evaluate

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