Solve 2t^2+24t+64

$$2 t ^ { 2 } + 24 t + 64$$

Factor

$$2\left(t+4\right)\left(t+8\right)$$

Solution Steps

Factor out 2.

$$2\left(t^{2}+12t+32\right)$$

Consider t^{2}+12t+32. Factor the expression by grouping. First, the expression needs to be rewritten as t^{2}+at+bt+32. To find a and b, set up a system to be solved.

$$a+b=12 ab=1\times 32=32$$

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 32.

$$1,32 2,16 4,8$$

Calculate the sum for each pair.

$$1+32=33 2+16=18 4+8=12$$

The solution is the pair that gives sum 12.

$$a=4 b=8$$

Rewrite t^{2}+12t+32 as \left(t^{2}+4t\right)+\left(8t+32\right).

$$\left(t^{2}+4t\right)+\left(8t+32\right)$$

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

$$t\left(t+4\right)+8\left(t+4\right)$$

Factor out common term t+4 by using distributive property.

$$\left(t+4\right)\left(t+8\right)$$

Rewrite the complete factored expression.

$$2\left(t+4\right)\left(t+8\right)$$

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.

$$2t^{2}+24t+64=0$$

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.

$$t=\frac{-24±\sqrt{24^{2}-4\times 2\times 64}}{2\times 2}$$

Square 24.

$$t=\frac{-24±\sqrt{576-4\times 2\times 64}}{2\times 2}$$

Multiply -4 times 2.

$$t=\frac{-24±\sqrt{576-8\times 64}}{2\times 2}$$

Multiply -8 times 64.

$$t=\frac{-24±\sqrt{576-512}}{2\times 2}$$

Add 576 to -512.

$$t=\frac{-24±\sqrt{64}}{2\times 2}$$

Take the square root of 64.

$$t=\frac{-24±8}{2\times 2}$$

Multiply 2 times 2.

$$t=\frac{-24±8}{4}$$

Now solve the equation t=\frac{-24±8}{4} when ± is plus. Add -24 to 8.

$$t=-\frac{16}{4}$$

Divide -16 by 4.

$$t=-4$$

Now solve the equation t=\frac{-24±8}{4} when ± is minus. Subtract 8 from -24.

$$t=-\frac{32}{4}$$

Divide -32 by 4.

$$t=-8$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -4 for x_{1} and -8 for x_{2}.

$$2t^{2}+24t+64=2\left(t-\left(-4\right)\right)\left(t-\left(-8\right)\right)$$

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

$$2t^{2}+24t+64=2\left(t+4\right)\left(t+8\right)$$

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

$$x ^ 2 +12x +32 = 0$$

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

$$r + s = -12 rs = 32$$

Two numbers r and s sum up to -12 exactly when the average of the two numbers is \frac{1}{2}*-12 = -6. 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>

$$r = -6 - u s = -6 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 32

$$(-6 - u) (-6 + u) = 32$$

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

$$36 - u^2 = 32$$

Simplify the expression by subtracting 36 on both sides

$$-u^2 = 32-36 = -4$$

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

$$u^2 = 4 u = \pm\sqrt{4} = \pm 2$$

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

$$r =-6 - 2 = -8 s = -6 + 2 = -4$$

Evaluate

$$2\left(t+4\right)\left(t+8\right)$$