Solve 2x^2+32x+30

$$2 x ^ { 2 } + 32 x + 30$$

Factor

$$2\left(x+1\right)\left(x+15\right)$$

Solution Steps

Factor out 2.

$$2\left(x^{2}+16x+15\right)$$

Consider x^{2}+16x+15. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+15. To find a and b, set up a system to be solved.

$$a+b=16 ab=1\times 15=15$$

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 15.

$$1,15 3,5$$

Calculate the sum for each pair.

$$1+15=16 3+5=8$$

The solution is the pair that gives sum 16.

$$a=1 b=15$$

Rewrite x^{2}+16x+15 as \left(x^{2}+x\right)+\left(15x+15\right).

$$\left(x^{2}+x\right)+\left(15x+15\right)$$

Factor out x in the first and 15 in the second group.

$$x\left(x+1\right)+15\left(x+1\right)$$

Factor out common term x+1 by using distributive property.

$$\left(x+1\right)\left(x+15\right)$$

Rewrite the complete factored expression.

$$2\left(x+1\right)\left(x+15\right)$$

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.

$$2x^{2}+32x+30=0$$

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.

$$x=\frac{-32±\sqrt{32^{2}-4\times 2\times 30}}{2\times 2}$$

Square 32.

$$x=\frac{-32±\sqrt{1024-4\times 2\times 30}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-32±\sqrt{1024-8\times 30}}{2\times 2}$$

Multiply -8 times 30.

$$x=\frac{-32±\sqrt{1024-240}}{2\times 2}$$

Add 1024 to -240.

$$x=\frac{-32±\sqrt{784}}{2\times 2}$$

Take the square root of 784.

$$x=\frac{-32±28}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-32±28}{4}$$

Now solve the equation x=\frac{-32±28}{4} when ± is plus. Add -32 to 28.

$$x=-\frac{4}{4}$$

Divide -4 by 4.

$$x=-1$$

Now solve the equation x=\frac{-32±28}{4} when ± is minus. Subtract 28 from -32.

$$x=-\frac{60}{4}$$

Divide -60 by 4.

$$x=-15$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -1 for x_{1} and -15 for x_{2}.

$$2x^{2}+32x+30=2\left(x-\left(-1\right)\right)\left(x-\left(-15\right)\right)$$

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

$$2x^{2}+32x+30=2\left(x+1\right)\left(x+15\right)$$

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.This is achieved by dividing both sides of the equation by 2

$$x ^ 2 +16x +15 = 0$$

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

$$r + s = -16 rs = 15$$

Two numbers r and s sum up to -16 exactly when the average of the two numbers is \frac{1}{2}*-16 = -8. 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>

$$r = -8 - u s = -8 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 15

$$(-8 - u) (-8 + u) = 15$$

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

$$64 - u^2 = 15$$

Simplify the expression by subtracting 64 on both sides

$$-u^2 = 15-64 = -49$$

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

$$u^2 = 49 u = \pm\sqrt{49} = \pm 7$$

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

$$r =-8 - 7 = -15 s = -8 + 7 = -1$$

Evaluate

$$2\left(x+1\right)\left(x+15\right)$$