Solve 12g^2-25g+12

$$12 g ^ { 2 } - 25 g + 12$$

Factor

$$\left(3g-4\right)\left(4g-3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 12g^{2}+ag+bg+12. To find a and b, set up a system to be solved.

$$a+b=-25 ab=12\times 12=144$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 144.

$$-1,-144 -2,-72 -3,-48 -4,-36 -6,-24 -8,-18 -9,-16 -12,-12$$

Calculate the sum for each pair.

$$-1-144=-145 -2-72=-74 -3-48=-51 -4-36=-40 -6-24=-30 -8-18=-26 -9-16=-25 -12-12=-24$$

The solution is the pair that gives sum -25.

$$a=-16 b=-9$$

Rewrite 12g^{2}-25g+12 as \left(12g^{2}-16g\right)+\left(-9g+12\right).

$$\left(12g^{2}-16g\right)+\left(-9g+12\right)$$

Factor out 4g in the first and -3 in the second group.

$$4g\left(3g-4\right)-3\left(3g-4\right)$$

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

$$\left(3g-4\right)\left(4g-3\right)$$

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.

$$12g^{2}-25g+12=0$$

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.

$$g=\frac{-\left(-25\right)±\sqrt{\left(-25\right)^{2}-4\times 12\times 12}}{2\times 12}$$

Square -25.

$$g=\frac{-\left(-25\right)±\sqrt{625-4\times 12\times 12}}{2\times 12}$$

Multiply -4 times 12.

$$g=\frac{-\left(-25\right)±\sqrt{625-48\times 12}}{2\times 12}$$

Multiply -48 times 12.

$$g=\frac{-\left(-25\right)±\sqrt{625-576}}{2\times 12}$$

Add 625 to -576.

$$g=\frac{-\left(-25\right)±\sqrt{49}}{2\times 12}$$

Take the square root of 49.

$$g=\frac{-\left(-25\right)±7}{2\times 12}$$

The opposite of -25 is 25.

$$g=\frac{25±7}{2\times 12}$$

Multiply 2 times 12.

$$g=\frac{25±7}{24}$$

Now solve the equation g=\frac{25±7}{24} when ± is plus. Add 25 to 7.

$$g=\frac{32}{24}$$

Reduce the fraction \frac{32}{24} to lowest terms by extracting and canceling out 8.

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

Now solve the equation g=\frac{25±7}{24} when ± is minus. Subtract 7 from 25.

$$g=\frac{18}{24}$$

Reduce the fraction \frac{18}{24} to lowest terms by extracting and canceling out 6.

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

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

$$12g^{2}-25g+12=12\left(g-\frac{4}{3}\right)\left(g-\frac{3}{4}\right)$$

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

$$12g^{2}-25g+12=12\times \frac{3g-4}{3}\left(g-\frac{3}{4}\right)$$

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

$$12g^{2}-25g+12=12\times \frac{3g-4}{3}\times \frac{4g-3}{4}$$

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

$$12g^{2}-25g+12=12\times \frac{\left(3g-4\right)\left(4g-3\right)}{3\times 4}$$

Multiply 3 times 4.

$$12g^{2}-25g+12=12\times \frac{\left(3g-4\right)\left(4g-3\right)}{12}$$

Cancel out 12, the greatest common factor in 12 and 12.

$$12g^{2}-25g+12=\left(3g-4\right)\left(4g-3\right)$$

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 12

$$x ^ 2 -\frac{25}{12}x +1 = 0$$

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

$$r + s = \frac{25}{12} rs = 1$$

Two numbers r and s sum up to \frac{25}{12} exactly when the average of the two numbers is \frac{1}{2}*\frac{25}{12} = \frac{25}{24}. 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>

$$r = \frac{25}{24} - u s = \frac{25}{24} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 1

$$(\frac{25}{24} - u) (\frac{25}{24} + u) = 1$$

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

$$\frac{625}{576} - u^2 = 1$$

Simplify the expression by subtracting \frac{625}{576} on both sides

$$-u^2 = 1-\frac{625}{576} = -\frac{49}{576}$$

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

$$u^2 = \frac{49}{576} u = \pm\sqrt{\frac{49}{576}} = \pm \frac{7}{24}$$

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

$$r =\frac{25}{24} - \frac{7}{24} = 0.750 s = \frac{25}{24} + \frac{7}{24} = 1.333$$

Evaluate

$$\left(3g-4\right)\left(4g-3\right)$$