Solve 3a^2-4a-20

$$3 a ^ { 2 } - 4 a - 20$$

Factor

$$\left(3a-10\right)\left(a+2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3a^{2}+pa+qa-20. To find p and q, set up a system to be solved.

$$p+q=-4 pq=3\left(-20\right)=-60$$

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 -60.

$$1,-60 2,-30 3,-20 4,-15 5,-12 6,-10$$

Calculate the sum for each pair.

$$1-60=-59 2-30=-28 3-20=-17 4-15=-11 5-12=-7 6-10=-4$$

The solution is the pair that gives sum -4.

$$p=-10 q=6$$

Rewrite 3a^{2}-4a-20 as \left(3a^{2}-10a\right)+\left(6a-20\right).

$$\left(3a^{2}-10a\right)+\left(6a-20\right)$$

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

$$a\left(3a-10\right)+2\left(3a-10\right)$$

Factor out common term 3a-10 by using distributive property.

$$\left(3a-10\right)\left(a+2\right)$$

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.

$$3a^{2}-4a-20=0$$

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.

$$a=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 3\left(-20\right)}}{2\times 3}$$

Square -4.

$$a=\frac{-\left(-4\right)±\sqrt{16-4\times 3\left(-20\right)}}{2\times 3}$$

Multiply -4 times 3.

$$a=\frac{-\left(-4\right)±\sqrt{16-12\left(-20\right)}}{2\times 3}$$

Multiply -12 times -20.

$$a=\frac{-\left(-4\right)±\sqrt{16+240}}{2\times 3}$$

Add 16 to 240.

$$a=\frac{-\left(-4\right)±\sqrt{256}}{2\times 3}$$

Take the square root of 256.

$$a=\frac{-\left(-4\right)±16}{2\times 3}$$

The opposite of -4 is 4.

$$a=\frac{4±16}{2\times 3}$$

Multiply 2 times 3.

$$a=\frac{4±16}{6}$$

Now solve the equation a=\frac{4±16}{6} when ± is plus. Add 4 to 16.

$$a=\frac{20}{6}$$

Reduce the fraction \frac{20}{6} to lowest terms by extracting and canceling out 2.

$$a=\frac{10}{3}$$

Now solve the equation a=\frac{4±16}{6} when ± is minus. Subtract 16 from 4.

$$a=-\frac{12}{6}$$

Divide -12 by 6.

$$a=-2$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{10}{3} for x_{1} and -2 for x_{2}.

$$3a^{2}-4a-20=3\left(a-\frac{10}{3}\right)\left(a-\left(-2\right)\right)$$

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

$$3a^{2}-4a-20=3\left(a-\frac{10}{3}\right)\left(a+2\right)$$

Subtract \frac{10}{3} from a by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$3a^{2}-4a-20=3\times \frac{3a-10}{3}\left(a+2\right)$$

Cancel out 3, the greatest common factor in 3 and 3.

$$3a^{2}-4a-20=\left(3a-10\right)\left(a+2\right)$$

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 3

$$x ^ 2 -\frac{4}{3}x -\frac{20}{3} = 0$$

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

$$r + s = \frac{4}{3} rs = -\frac{20}{3}$$

Two numbers r and s sum up to \frac{4}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{4}{3} = \frac{2}{3}. 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>

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

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{20}{3}

$$(\frac{2}{3} - u) (\frac{2}{3} + u) = -\frac{20}{3}$$

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

$$\frac{4}{9} - u^2 = -\frac{20}{3}$$

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

$$-u^2 = -\frac{20}{3}-\frac{4}{9} = -\frac{64}{9}$$

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

$$u^2 = \frac{64}{9} u = \pm\sqrt{\frac{64}{9}} = \pm \frac{8}{3}$$

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

$$r =\frac{2}{3} - \frac{8}{3} = -2.000 s = \frac{2}{3} + \frac{8}{3} = 3.333$$

Evaluate

$$\left(3a-10\right)\left(a+2\right)$$