Solve 12a^2-36a+15

$$12 a ^ { 2 } - 36 a + 15$$

Factor

$$3\left(2a-5\right)\left(2a-1\right)$$

Solution Steps

Factor out 3.

$$3\left(4a^{2}-12a+5\right)$$

Consider 4a^{2}-12a+5. Factor the expression by grouping. First, the expression needs to be rewritten as 4a^{2}+pa+qa+5. To find p and q, set up a system to be solved.

$$p+q=-12 pq=4\times 5=20$$

Since pq is positive, p and q have the same sign. Since p+q is negative, p and q are both negative. List all such integer pairs that give product 20.

$$-1,-20 -2,-10 -4,-5$$

Calculate the sum for each pair.

$$-1-20=-21 -2-10=-12 -4-5=-9$$

The solution is the pair that gives sum -12.

$$p=-10 q=-2$$

Rewrite 4a^{2}-12a+5 as \left(4a^{2}-10a\right)+\left(-2a+5\right).

$$\left(4a^{2}-10a\right)+\left(-2a+5\right)$$

Factor out 2a in the first and -1 in the second group.

$$2a\left(2a-5\right)-\left(2a-5\right)$$

Factor out common term 2a-5 by using distributive property.

$$\left(2a-5\right)\left(2a-1\right)$$

Rewrite the complete factored expression.

$$3\left(2a-5\right)\left(2a-1\right)$$

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.

$$12a^{2}-36a+15=0$$

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.

$$a=\frac{-\left(-36\right)±\sqrt{\left(-36\right)^{2}-4\times 12\times 15}}{2\times 12}$$

Square -36.

$$a=\frac{-\left(-36\right)±\sqrt{1296-4\times 12\times 15}}{2\times 12}$$

Multiply -4 times 12.

$$a=\frac{-\left(-36\right)±\sqrt{1296-48\times 15}}{2\times 12}$$

Multiply -48 times 15.

$$a=\frac{-\left(-36\right)±\sqrt{1296-720}}{2\times 12}$$

Add 1296 to -720.

$$a=\frac{-\left(-36\right)±\sqrt{576}}{2\times 12}$$

Take the square root of 576.

$$a=\frac{-\left(-36\right)±24}{2\times 12}$$

The opposite of -36 is 36.

$$a=\frac{36±24}{2\times 12}$$

Multiply 2 times 12.

$$a=\frac{36±24}{24}$$

Now solve the equation a=\frac{36±24}{24} when ± is plus. Add 36 to 24.

$$a=\frac{60}{24}$$

Reduce the fraction \frac{60}{24} to lowest terms by extracting and canceling out 12.

$$a=\frac{5}{2}$$

Now solve the equation a=\frac{36±24}{24} when ± is minus. Subtract 24 from 36.

$$a=\frac{12}{24}$$

Reduce the fraction \frac{12}{24} to lowest terms by extracting and canceling out 12.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{2} for x_{1} and \frac{1}{2} for x_{2}.

$$12a^{2}-36a+15=12\left(a-\frac{5}{2}\right)\left(a-\frac{1}{2}\right)$$

Subtract \frac{5}{2} from a by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$12a^{2}-36a+15=12\times \frac{2a-5}{2}\left(a-\frac{1}{2}\right)$$

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

$$12a^{2}-36a+15=12\times \frac{2a-5}{2}\times \frac{2a-1}{2}$$

Multiply \frac{2a-5}{2} times \frac{2a-1}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$12a^{2}-36a+15=12\times \frac{\left(2a-5\right)\left(2a-1\right)}{2\times 2}$$

Multiply 2 times 2.

$$12a^{2}-36a+15=12\times \frac{\left(2a-5\right)\left(2a-1\right)}{4}$$

Cancel out 4, the greatest common factor in 12 and 4.

$$12a^{2}-36a+15=3\left(2a-5\right)\left(2a-1\right)$$

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

$$x ^ 2 -3x +\frac{5}{4} = 0$$

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

$$r + s = 3 rs = \frac{5}{4}$$

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

$$r = \frac{3}{2} - u s = \frac{3}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{5}{4}

$$(\frac{3}{2} - u) (\frac{3}{2} + u) = \frac{5}{4}$$

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

$$\frac{9}{4} - u^2 = \frac{5}{4}$$

Simplify the expression by subtracting \frac{9}{4} on both sides

$$-u^2 = \frac{5}{4}-\frac{9}{4} = -1$$

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

$$u^2 = 1 u = \pm\sqrt{1} = \pm 1$$

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

$$r =\frac{3}{2} - 1 = 0.500 s = \frac{3}{2} + 1 = 2.500$$

Evaluate

$$12a^{2}-36a+15$$