Solve 4t^2-13t-12

$$4 t ^ { 2 } - 13 t - 12$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 4t^{2}+at+bt-12. To find a and b, set up a system to be solved.

$$a+b=-13 ab=4\left(-12\right)=-48$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -48.

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

Calculate the sum for each pair.

$$1-48=-47 2-24=-22 3-16=-13 4-12=-8 6-8=-2$$

The solution is the pair that gives sum -13.

$$a=-16 b=3$$

Rewrite 4t^{2}-13t-12 as \left(4t^{2}-16t\right)+\left(3t-12\right).

$$\left(4t^{2}-16t\right)+\left(3t-12\right)$$

Factor out 4t in the first and 3 in the second group.

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

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

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

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.

$$4t^{2}-13t-12=0$$

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.

$$t=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 4\left(-12\right)}}{2\times 4}$$

Square -13.

$$t=\frac{-\left(-13\right)±\sqrt{169-4\times 4\left(-12\right)}}{2\times 4}$$

Multiply -4 times 4.

$$t=\frac{-\left(-13\right)±\sqrt{169-16\left(-12\right)}}{2\times 4}$$

Multiply -16 times -12.

$$t=\frac{-\left(-13\right)±\sqrt{169+192}}{2\times 4}$$

Add 169 to 192.

$$t=\frac{-\left(-13\right)±\sqrt{361}}{2\times 4}$$

Take the square root of 361.

$$t=\frac{-\left(-13\right)±19}{2\times 4}$$

The opposite of -13 is 13.

$$t=\frac{13±19}{2\times 4}$$

Multiply 2 times 4.

$$t=\frac{13±19}{8}$$

Now solve the equation t=\frac{13±19}{8} when ± is plus. Add 13 to 19.

$$t=\frac{32}{8}$$

Divide 32 by 8.

$$t=4$$

Now solve the equation t=\frac{13±19}{8} when ± is minus. Subtract 19 from 13.

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

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

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

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 -\frac{3}{4} for x_{2}.

$$4t^{2}-13t-12=4\left(t-4\right)\left(t-\left(-\frac{3}{4}\right)\right)$$

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

$$4t^{2}-13t-12=4\left(t-4\right)\left(t+\frac{3}{4}\right)$$

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

$$4t^{2}-13t-12=4\left(t-4\right)\times \frac{4t+3}{4}$$

Cancel out 4, the greatest common factor in 4 and 4.

$$4t^{2}-13t-12=\left(t-4\right)\left(4t+3\right)$$

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 4

$$x ^ 2 -\frac{13}{4}x -3 = 0$$

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

$$r + s = \frac{13}{4} rs = -3$$

Two numbers r and s sum up to \frac{13}{4} exactly when the average of the two numbers is \frac{1}{2}*\frac{13}{4} = \frac{13}{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>

$$r = \frac{13}{8} - u s = \frac{13}{8} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -3

$$(\frac{13}{8} - u) (\frac{13}{8} + u) = -3$$

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

$$\frac{169}{64} - u^2 = -3$$

Simplify the expression by subtracting \frac{169}{64} on both sides

$$-u^2 = -3-\frac{169}{64} = -\frac{361}{64}$$

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

$$u^2 = \frac{361}{64} u = \pm\sqrt{\frac{361}{64}} = \pm \frac{19}{8}$$

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

$$r =\frac{13}{8} - \frac{19}{8} = -0.750 s = \frac{13}{8} + \frac{19}{8} = 4$$

Evaluate

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