Solve 5t^2+28t+32

$$5 t ^ { 2 } + 28 t + 32$$

Factor

$$\left(t+4\right)\left(5t+8\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 5t^{2}+at+bt+32. To find a and b, set up a system to be solved.

$$a+b=28 ab=5\times 32=160$$

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 160.

$$1,160 2,80 4,40 5,32 8,20 10,16$$

Calculate the sum for each pair.

$$1+160=161 2+80=82 4+40=44 5+32=37 8+20=28 10+16=26$$

The solution is the pair that gives sum 28.

$$a=8 b=20$$

Rewrite 5t^{2}+28t+32 as \left(5t^{2}+8t\right)+\left(20t+32\right).

$$\left(5t^{2}+8t\right)+\left(20t+32\right)$$

Factor out t in the first and 4 in the second group.

$$t\left(5t+8\right)+4\left(5t+8\right)$$

Factor out common term 5t+8 by using distributive property.

$$\left(5t+8\right)\left(t+4\right)$$

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.

$$5t^{2}+28t+32=0$$

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.

$$t=\frac{-28±\sqrt{28^{2}-4\times 5\times 32}}{2\times 5}$$

Square 28.

$$t=\frac{-28±\sqrt{784-4\times 5\times 32}}{2\times 5}$$

Multiply -4 times 5.

$$t=\frac{-28±\sqrt{784-20\times 32}}{2\times 5}$$

Multiply -20 times 32.

$$t=\frac{-28±\sqrt{784-640}}{2\times 5}$$

Add 784 to -640.

$$t=\frac{-28±\sqrt{144}}{2\times 5}$$

Take the square root of 144.

$$t=\frac{-28±12}{2\times 5}$$

Multiply 2 times 5.

$$t=\frac{-28±12}{10}$$

Now solve the equation t=\frac{-28±12}{10} when ± is plus. Add -28 to 12.

$$t=-\frac{16}{10}$$

Reduce the fraction \frac{-16}{10} to lowest terms by extracting and canceling out 2.

$$t=-\frac{8}{5}$$

Now solve the equation t=\frac{-28±12}{10} when ± is minus. Subtract 12 from -28.

$$t=-\frac{40}{10}$$

Divide -40 by 10.

$$t=-4$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{8}{5} for x_{1} and -4 for x_{2}.

$$5t^{2}+28t+32=5\left(t-\left(-\frac{8}{5}\right)\right)\left(t-\left(-4\right)\right)$$

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

$$5t^{2}+28t+32=5\left(t+\frac{8}{5}\right)\left(t+4\right)$$

Add \frac{8}{5} to t by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$5t^{2}+28t+32=5\times \frac{5t+8}{5}\left(t+4\right)$$

Cancel out 5, the greatest common factor in 5 and 5.

$$5t^{2}+28t+32=\left(5t+8\right)\left(t+4\right)$$

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 5

$$x ^ 2 +\frac{28}{5}x +\frac{32}{5} = 0$$

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

$$r + s = -\frac{28}{5} rs = \frac{32}{5}$$

Two numbers r and s sum up to -\frac{28}{5} exactly when the average of the two numbers is \frac{1}{2}*-\frac{28}{5} = -\frac{14}{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>

$$r = -\frac{14}{5} - u s = -\frac{14}{5} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{32}{5}

$$(-\frac{14}{5} - u) (-\frac{14}{5} + u) = \frac{32}{5}$$

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

$$\frac{196}{25} - u^2 = \frac{32}{5}$$

Simplify the expression by subtracting \frac{196}{25} on both sides

$$-u^2 = \frac{32}{5}-\frac{196}{25} = -\frac{36}{25}$$

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

$$u^2 = \frac{36}{25} u = \pm\sqrt{\frac{36}{25}} = \pm \frac{6}{5}$$

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

$$r =-\frac{14}{5} - \frac{6}{5} = -4.000 s = -\frac{14}{5} + \frac{6}{5} = -1.600$$

Evaluate

$$\left(t+4\right)\left(5t+8\right)$$