Solve 12a^2-10a-42

$$12 a ^ { 2 } - 10 a - 42$$

Factor

$$2\left(3a-7\right)\left(2a+3\right)$$

Solution Steps

Factor out 2.

$$2\left(6a^{2}-5a-21\right)$$

Consider 6a^{2}-5a-21. Factor the expression by grouping. First, the expression needs to be rewritten as 6a^{2}+pa+qa-21. To find p and q, set up a system to be solved.

$$p+q=-5 pq=6\left(-21\right)=-126$$

Since pq is negative, p and q have the opposite signs. Since p+q is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -126.

$$1,-126 2,-63 3,-42 6,-21 7,-18 9,-14$$

Calculate the sum for each pair.

$$1-126=-125 2-63=-61 3-42=-39 6-21=-15 7-18=-11 9-14=-5$$

The solution is the pair that gives sum -5.

$$p=-14 q=9$$

Rewrite 6a^{2}-5a-21 as \left(6a^{2}-14a\right)+\left(9a-21\right).

$$\left(6a^{2}-14a\right)+\left(9a-21\right)$$

Factor out 2a in the first and 3 in the second group.

$$2a\left(3a-7\right)+3\left(3a-7\right)$$

Factor out common term 3a-7 by using distributive property.

$$\left(3a-7\right)\left(2a+3\right)$$

Rewrite the complete factored expression.

$$2\left(3a-7\right)\left(2a+3\right)$$

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.

$$12a^{2}-10a-42=0$$

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.

$$a=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 12\left(-42\right)}}{2\times 12}$$

Square -10.

$$a=\frac{-\left(-10\right)±\sqrt{100-4\times 12\left(-42\right)}}{2\times 12}$$

Multiply -4 times 12.

$$a=\frac{-\left(-10\right)±\sqrt{100-48\left(-42\right)}}{2\times 12}$$

Multiply -48 times -42.

$$a=\frac{-\left(-10\right)±\sqrt{100+2016}}{2\times 12}$$

Add 100 to 2016.

$$a=\frac{-\left(-10\right)±\sqrt{2116}}{2\times 12}$$

Take the square root of 2116.

$$a=\frac{-\left(-10\right)±46}{2\times 12}$$

The opposite of -10 is 10.

$$a=\frac{10±46}{2\times 12}$$

Multiply 2 times 12.

$$a=\frac{10±46}{24}$$

Now solve the equation a=\frac{10±46}{24} when ± is plus. Add 10 to 46.

$$a=\frac{56}{24}$$

Reduce the fraction \frac{56}{24} to lowest terms by extracting and canceling out 8.

$$a=\frac{7}{3}$$

Now solve the equation a=\frac{10±46}{24} when ± is minus. Subtract 46 from 10.

$$a=-\frac{36}{24}$$

Reduce the fraction \frac{-36}{24} to lowest terms by extracting and canceling out 12.

$$a=-\frac{3}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{7}{3} for x_{1} and -\frac{3}{2} for x_{2}.

$$12a^{2}-10a-42=12\left(a-\frac{7}{3}\right)\left(a-\left(-\frac{3}{2}\right)\right)$$

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

$$12a^{2}-10a-42=12\left(a-\frac{7}{3}\right)\left(a+\frac{3}{2}\right)$$

Subtract \frac{7}{3} from a by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$12a^{2}-10a-42=12\times \frac{3a-7}{3}\left(a+\frac{3}{2}\right)$$

Add \frac{3}{2} to a by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$12a^{2}-10a-42=12\times \frac{3a-7}{3}\times \frac{2a+3}{2}$$

Multiply \frac{3a-7}{3} times \frac{2a+3}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$12a^{2}-10a-42=12\times \frac{\left(3a-7\right)\left(2a+3\right)}{3\times 2}$$

Multiply 3 times 2.

$$12a^{2}-10a-42=12\times \frac{\left(3a-7\right)\left(2a+3\right)}{6}$$

Cancel out 6, the greatest common factor in 12 and 6.

$$12a^{2}-10a-42=2\left(3a-7\right)\left(2a+3\right)$$

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

$$x ^ 2 -\frac{5}{6}x -\frac{7}{2} = 0$$

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

$$r + s = \frac{5}{6} rs = -\frac{7}{2}$$

Two numbers r and s sum up to \frac{5}{6} exactly when the average of the two numbers is \frac{1}{2}*\frac{5}{6} = \frac{5}{12}. 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>

$$r = \frac{5}{12} - u s = \frac{5}{12} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{7}{2}

$$(\frac{5}{12} - u) (\frac{5}{12} + u) = -\frac{7}{2}$$

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

$$\frac{25}{144} - u^2 = -\frac{7}{2}$$

Simplify the expression by subtracting \frac{25}{144} on both sides

$$-u^2 = -\frac{7}{2}-\frac{25}{144} = -\frac{529}{144}$$

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

$$u^2 = \frac{529}{144} u = \pm\sqrt{\frac{529}{144}} = \pm \frac{23}{12}$$

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

$$r =\frac{5}{12} - \frac{23}{12} = -1.500 s = \frac{5}{12} + \frac{23}{12} = 2.333$$

Evaluate

$$12a^{2}-10a-42$$