Solve 5q^2-17q-12

$$5 q ^ { 2 } - 17 q - 12$$

Factor

$$\left(q-4\right)\left(5q+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 5q^{2}+aq+bq-12. To find a and b, set up a system to be solved.

$$a+b=-17 ab=5\left(-12\right)=-60$$

Since ab is negative, a and b have the opposite signs. Since a+b 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 -17.

$$a=-20 b=3$$

Rewrite 5q^{2}-17q-12 as \left(5q^{2}-20q\right)+\left(3q-12\right).

$$\left(5q^{2}-20q\right)+\left(3q-12\right)$$

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

$$5q\left(q-4\right)+3\left(q-4\right)$$

Factor out common term q-4 by using distributive property.

$$\left(q-4\right)\left(5q+3\right)$$

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.

$$5q^{2}-17q-12=0$$

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.

$$q=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 5\left(-12\right)}}{2\times 5}$$

Square -17.

$$q=\frac{-\left(-17\right)±\sqrt{289-4\times 5\left(-12\right)}}{2\times 5}$$

Multiply -4 times 5.

$$q=\frac{-\left(-17\right)±\sqrt{289-20\left(-12\right)}}{2\times 5}$$

Multiply -20 times -12.

$$q=\frac{-\left(-17\right)±\sqrt{289+240}}{2\times 5}$$

Add 289 to 240.

$$q=\frac{-\left(-17\right)±\sqrt{529}}{2\times 5}$$

Take the square root of 529.

$$q=\frac{-\left(-17\right)±23}{2\times 5}$$

The opposite of -17 is 17.

$$q=\frac{17±23}{2\times 5}$$

Multiply 2 times 5.

$$q=\frac{17±23}{10}$$

Now solve the equation q=\frac{17±23}{10} when ± is plus. Add 17 to 23.

$$q=\frac{40}{10}$$

Divide 40 by 10.

$$q=4$$

Now solve the equation q=\frac{17±23}{10} when ± is minus. Subtract 23 from 17.

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

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

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 4 for x_{1} and -\frac{3}{5} for x_{2}.

$$5q^{2}-17q-12=5\left(q-4\right)\left(q-\left(-\frac{3}{5}\right)\right)$$

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

$$5q^{2}-17q-12=5\left(q-4\right)\left(q+\frac{3}{5}\right)$$

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

$$5q^{2}-17q-12=5\left(q-4\right)\times \frac{5q+3}{5}$$

Cancel out 5, the greatest common factor in 5 and 5.

$$5q^{2}-17q-12=\left(q-4\right)\left(5q+3\right)$$

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 5

$$x ^ 2 -\frac{17}{5}x -\frac{12}{5} = 0$$

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

$$r + s = \frac{17}{5} rs = -\frac{12}{5}$$

Two numbers r and s sum up to \frac{17}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{17}{5} = \frac{17}{10}. 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>

$$r = \frac{17}{10} - u s = \frac{17}{10} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{12}{5}

$$(\frac{17}{10} - u) (\frac{17}{10} + u) = -\frac{12}{5}$$

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

$$\frac{289}{100} - u^2 = -\frac{12}{5}$$

Simplify the expression by subtracting \frac{289}{100} on both sides

$$-u^2 = -\frac{12}{5}-\frac{289}{100} = -\frac{529}{100}$$

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

$$u^2 = \frac{529}{100} u = \pm\sqrt{\frac{529}{100}} = \pm \frac{23}{10}$$

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

$$r =\frac{17}{10} - \frac{23}{10} = -0.600 s = \frac{17}{10} + \frac{23}{10} = 4$$

Evaluate

$$\left(q-4\right)\left(5q+3\right)$$