Solve 4t^2+13t-12

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

Factor

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

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 positive, the positive number has greater absolute value than the negative. 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.

$$a=-3 b=16$$

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

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

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

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

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

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

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.

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

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.

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

Square 13.

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

Multiply -4 times 4.

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

Multiply -16 times -12.

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

Add 169 to 192.

$$t=\frac{-13±\sqrt{361}}{2\times 4}$$

Take the square root of 361.

$$t=\frac{-13±19}{2\times 4}$$

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.

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

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

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

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

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

Divide -32 by 8.

$$t=-4$$

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

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

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

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

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

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

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

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

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 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.

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

Evaluate

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