Solve 3t^2+20t-32

$$3 t ^ { 2 } + 20 t - 32$$

Factor

$$\left(3t-4\right)\left(t+8\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3t^{2}+at+bt-32. To find a and b, set up a system to be solved.

$$a+b=20 ab=3\left(-32\right)=-96$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -96.

$$-1,96 -2,48 -3,32 -4,24 -6,16 -8,12$$

Calculate the sum for each pair.

$$-1+96=95 -2+48=46 -3+32=29 -4+24=20 -6+16=10 -8+12=4$$

The solution is the pair that gives sum 20.

$$a=-4 b=24$$

Rewrite 3t^{2}+20t-32 as \left(3t^{2}-4t\right)+\left(24t-32\right).

$$\left(3t^{2}-4t\right)+\left(24t-32\right)$$

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

$$t\left(3t-4\right)+8\left(3t-4\right)$$

Factor out common term 3t-4 by using distributive property.

$$\left(3t-4\right)\left(t+8\right)$$

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.

$$3t^{2}+20t-32=0$$

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.

$$t=\frac{-20±\sqrt{20^{2}-4\times 3\left(-32\right)}}{2\times 3}$$

Square 20.

$$t=\frac{-20±\sqrt{400-4\times 3\left(-32\right)}}{2\times 3}$$

Multiply -4 times 3.

$$t=\frac{-20±\sqrt{400-12\left(-32\right)}}{2\times 3}$$

Multiply -12 times -32.

$$t=\frac{-20±\sqrt{400+384}}{2\times 3}$$

Add 400 to 384.

$$t=\frac{-20±\sqrt{784}}{2\times 3}$$

Take the square root of 784.

$$t=\frac{-20±28}{2\times 3}$$

Multiply 2 times 3.

$$t=\frac{-20±28}{6}$$

Now solve the equation t=\frac{-20±28}{6} when ± is plus. Add -20 to 28.

$$t=\frac{8}{6}$$

Reduce the fraction \frac{8}{6} to lowest terms by extracting and canceling out 2.

$$t=\frac{4}{3}$$

Now solve the equation t=\frac{-20±28}{6} when ± is minus. Subtract 28 from -20.

$$t=-\frac{48}{6}$$

Divide -48 by 6.

$$t=-8$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{4}{3} for x_{1} and -8 for x_{2}.

$$3t^{2}+20t-32=3\left(t-\frac{4}{3}\right)\left(t-\left(-8\right)\right)$$

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

$$3t^{2}+20t-32=3\left(t-\frac{4}{3}\right)\left(t+8\right)$$

Subtract \frac{4}{3} from t by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$3t^{2}+20t-32=3\times \frac{3t-4}{3}\left(t+8\right)$$

Cancel out 3, the greatest common factor in 3 and 3.

$$3t^{2}+20t-32=\left(3t-4\right)\left(t+8\right)$$

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 3

$$x ^ 2 +\frac{20}{3}x -\frac{32}{3} = 0$$

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

$$r + s = -\frac{20}{3} rs = -\frac{32}{3}$$

Two numbers r and s sum up to -\frac{20}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{20}{3} = -\frac{10}{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>

$$r = -\frac{10}{3} - u s = -\frac{10}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{32}{3}

$$(-\frac{10}{3} - u) (-\frac{10}{3} + u) = -\frac{32}{3}$$

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

$$\frac{100}{9} - u^2 = -\frac{32}{3}$$

Simplify the expression by subtracting \frac{100}{9} on both sides

$$-u^2 = -\frac{32}{3}-\frac{100}{9} = -\frac{196}{9}$$

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

$$u^2 = \frac{196}{9} u = \pm\sqrt{\frac{196}{9}} = \pm \frac{14}{3}$$

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

$$r =-\frac{10}{3} - \frac{14}{3} = -8 s = -\frac{10}{3} + \frac{14}{3} = 1.333$$

Evaluate

$$\left(3t-4\right)\left(t+8\right)$$