Solve 3a^2-13a-30=

$$3 a ^ { 2 } - 13 a - 30 =$$

Factor

$$\left(a-6\right)\left(3a+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3a^{2}+pa+qa-30. To find p and q, set up a system to be solved.

$$p+q=-13 pq=3\left(-30\right)=-90$$

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 -90.

$$1,-90 2,-45 3,-30 5,-18 6,-15 9,-10$$

Calculate the sum for each pair.

$$1-90=-89 2-45=-43 3-30=-27 5-18=-13 6-15=-9 9-10=-1$$

The solution is the pair that gives sum -13.

$$p=-18 q=5$$

Rewrite 3a^{2}-13a-30 as \left(3a^{2}-18a\right)+\left(5a-30\right).

$$\left(3a^{2}-18a\right)+\left(5a-30\right)$$

Factor out 3a in the first and 5 in the second group.

$$3a\left(a-6\right)+5\left(a-6\right)$$

Factor out common term a-6 by using distributive property.

$$\left(a-6\right)\left(3a+5\right)$$

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.

$$3a^{2}-13a-30=0$$

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.

$$a=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 3\left(-30\right)}}{2\times 3}$$

Square -13.

$$a=\frac{-\left(-13\right)±\sqrt{169-4\times 3\left(-30\right)}}{2\times 3}$$

Multiply -4 times 3.

$$a=\frac{-\left(-13\right)±\sqrt{169-12\left(-30\right)}}{2\times 3}$$

Multiply -12 times -30.

$$a=\frac{-\left(-13\right)±\sqrt{169+360}}{2\times 3}$$

Add 169 to 360.

$$a=\frac{-\left(-13\right)±\sqrt{529}}{2\times 3}$$

Take the square root of 529.

$$a=\frac{-\left(-13\right)±23}{2\times 3}$$

The opposite of -13 is 13.

$$a=\frac{13±23}{2\times 3}$$

Multiply 2 times 3.

$$a=\frac{13±23}{6}$$

Now solve the equation a=\frac{13±23}{6} when ± is plus. Add 13 to 23.

$$a=\frac{36}{6}$$

Divide 36 by 6.

$$a=6$$

Now solve the equation a=\frac{13±23}{6} when ± is minus. Subtract 23 from 13.

$$a=-\frac{10}{6}$$

Reduce the fraction \frac{-10}{6} to lowest terms by extracting and canceling out 2.

$$a=-\frac{5}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 6 for x_{1} and -\frac{5}{3} for x_{2}.

$$3a^{2}-13a-30=3\left(a-6\right)\left(a-\left(-\frac{5}{3}\right)\right)$$

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

$$3a^{2}-13a-30=3\left(a-6\right)\left(a+\frac{5}{3}\right)$$

Add \frac{5}{3} to a by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$3a^{2}-13a-30=3\left(a-6\right)\times \frac{3a+5}{3}$$

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

$$3a^{2}-13a-30=\left(a-6\right)\left(3a+5\right)$$

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

$$x ^ 2 -\frac{13}{3}x -10 = 0$$

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

$$r + s = \frac{13}{3} rs = -10$$

Two numbers r and s sum up to \frac{13}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{13}{3} = \frac{13}{6}. 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>

$$r = \frac{13}{6} - u s = \frac{13}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -10

$$(\frac{13}{6} - u) (\frac{13}{6} + u) = -10$$

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

$$\frac{169}{36} - u^2 = -10$$

Simplify the expression by subtracting \frac{169}{36} on both sides

$$-u^2 = -10-\frac{169}{36} = -\frac{529}{36}$$

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

$$u^2 = \frac{529}{36} u = \pm\sqrt{\frac{529}{36}} = \pm \frac{23}{6}$$

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

$$r =\frac{13}{6} - \frac{23}{6} = -1.667 s = \frac{13}{6} + \frac{23}{6} = 6$$

Evaluate

$$\left(a-6\right)\left(3a+5\right)$$