Solve 8t^2+2t-3

$$8 t ^ { 2 } + 2 t - 3$$

Factor

$$\left(2t-1\right)\left(4t+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 8t^{2}+at+bt-3. To find a and b, set up a system to be solved.

$$a+b=2 ab=8\left(-3\right)=-24$$

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 -24.

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

Calculate the sum for each pair.

$$-1+24=23 -2+12=10 -3+8=5 -4+6=2$$

The solution is the pair that gives sum 2.

$$a=-4 b=6$$

Rewrite 8t^{2}+2t-3 as \left(8t^{2}-4t\right)+\left(6t-3\right).

$$\left(8t^{2}-4t\right)+\left(6t-3\right)$$

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

$$4t\left(2t-1\right)+3\left(2t-1\right)$$

Factor out common term 2t-1 by using distributive property.

$$\left(2t-1\right)\left(4t+3\right)$$

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.

$$8t^{2}+2t-3=0$$

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.

$$t=\frac{-2±\sqrt{2^{2}-4\times 8\left(-3\right)}}{2\times 8}$$

Square 2.

$$t=\frac{-2±\sqrt{4-4\times 8\left(-3\right)}}{2\times 8}$$

Multiply -4 times 8.

$$t=\frac{-2±\sqrt{4-32\left(-3\right)}}{2\times 8}$$

Multiply -32 times -3.

$$t=\frac{-2±\sqrt{4+96}}{2\times 8}$$

Add 4 to 96.

$$t=\frac{-2±\sqrt{100}}{2\times 8}$$

Take the square root of 100.

$$t=\frac{-2±10}{2\times 8}$$

Multiply 2 times 8.

$$t=\frac{-2±10}{16}$$

Now solve the equation t=\frac{-2±10}{16} when ± is plus. Add -2 to 10.

$$t=\frac{8}{16}$$

Reduce the fraction \frac{8}{16} to lowest terms by extracting and canceling out 8.

$$t=\frac{1}{2}$$

Now solve the equation t=\frac{-2±10}{16} when ± is minus. Subtract 10 from -2.

$$t=-\frac{12}{16}$$

Reduce the fraction \frac{-12}{16} to lowest terms by extracting and canceling out 4.

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

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

$$8t^{2}+2t-3=8\left(t-\frac{1}{2}\right)\left(t-\left(-\frac{3}{4}\right)\right)$$

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

$$8t^{2}+2t-3=8\left(t-\frac{1}{2}\right)\left(t+\frac{3}{4}\right)$$

Subtract \frac{1}{2} from t by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$8t^{2}+2t-3=8\times \frac{2t-1}{2}\left(t+\frac{3}{4}\right)$$

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

$$8t^{2}+2t-3=8\times \frac{2t-1}{2}\times \frac{4t+3}{4}$$

Multiply \frac{2t-1}{2} times \frac{4t+3}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$8t^{2}+2t-3=8\times \frac{\left(2t-1\right)\left(4t+3\right)}{2\times 4}$$

Multiply 2 times 4.

$$8t^{2}+2t-3=8\times \frac{\left(2t-1\right)\left(4t+3\right)}{8}$$

Cancel out 8, the greatest common factor in 8 and 8.

$$8t^{2}+2t-3=\left(2t-1\right)\left(4t+3\right)$$

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 8

$$x ^ 2 +\frac{1}{4}x -\frac{3}{8} = 0$$

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

$$r + s = -\frac{1}{4} rs = -\frac{3}{8}$$

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

$$r = -\frac{1}{8} - u s = -\frac{1}{8} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{8}

$$(-\frac{1}{8} - u) (-\frac{1}{8} + u) = -\frac{3}{8}$$

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

$$\frac{1}{64} - u^2 = -\frac{3}{8}$$

Simplify the expression by subtracting \frac{1}{64} on both sides

$$-u^2 = -\frac{3}{8}-\frac{1}{64} = -\frac{25}{64}$$

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

$$u^2 = \frac{25}{64} u = \pm\sqrt{\frac{25}{64}} = \pm \frac{5}{8}$$

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

$$r =-\frac{1}{8} - \frac{5}{8} = -0.750 s = -\frac{1}{8} + \frac{5}{8} = 0.500$$

Evaluate

$$\left(2t-1\right)\left(4t+3\right)$$