Solve 5j^2-12j-9

$$5 j ^ { 2 } - 12 j - 9$$

Factor

$$\left(j-3\right)\left(5j+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 5j^{2}+aj+bj-9. To find a and b, set up a system to be solved.

$$a+b=-12 ab=5\left(-9\right)=-45$$

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 -45.

$$1,-45 3,-15 5,-9$$

Calculate the sum for each pair.

$$1-45=-44 3-15=-12 5-9=-4$$

The solution is the pair that gives sum -12.

$$a=-15 b=3$$

Rewrite 5j^{2}-12j-9 as \left(5j^{2}-15j\right)+\left(3j-9\right).

$$\left(5j^{2}-15j\right)+\left(3j-9\right)$$

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

$$5j\left(j-3\right)+3\left(j-3\right)$$

Factor out common term j-3 by using distributive property.

$$\left(j-3\right)\left(5j+3\right)$$

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.

$$5j^{2}-12j-9=0$$

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.

$$j=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 5\left(-9\right)}}{2\times 5}$$

Square -12.

$$j=\frac{-\left(-12\right)±\sqrt{144-4\times 5\left(-9\right)}}{2\times 5}$$

Multiply -4 times 5.

$$j=\frac{-\left(-12\right)±\sqrt{144-20\left(-9\right)}}{2\times 5}$$

Multiply -20 times -9.

$$j=\frac{-\left(-12\right)±\sqrt{144+180}}{2\times 5}$$

Add 144 to 180.

$$j=\frac{-\left(-12\right)±\sqrt{324}}{2\times 5}$$

Take the square root of 324.

$$j=\frac{-\left(-12\right)±18}{2\times 5}$$

The opposite of -12 is 12.

$$j=\frac{12±18}{2\times 5}$$

Multiply 2 times 5.

$$j=\frac{12±18}{10}$$

Now solve the equation j=\frac{12±18}{10} when ± is plus. Add 12 to 18.

$$j=\frac{30}{10}$$

Divide 30 by 10.

$$j=3$$

Now solve the equation j=\frac{12±18}{10} when ± is minus. Subtract 18 from 12.

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

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

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

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

$$5j^{2}-12j-9=5\left(j-3\right)\left(j-\left(-\frac{3}{5}\right)\right)$$

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

$$5j^{2}-12j-9=5\left(j-3\right)\left(j+\frac{3}{5}\right)$$

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

$$5j^{2}-12j-9=5\left(j-3\right)\times \frac{5j+3}{5}$$

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

$$5j^{2}-12j-9=\left(j-3\right)\left(5j+3\right)$$

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

$$x ^ 2 -\frac{12}{5}x -\frac{9}{5} = 0$$

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

$$r + s = \frac{12}{5} rs = -\frac{9}{5}$$

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

$$r = \frac{6}{5} - u s = \frac{6}{5} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{9}{5}

$$(\frac{6}{5} - u) (\frac{6}{5} + u) = -\frac{9}{5}$$

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

$$\frac{36}{25} - u^2 = -\frac{9}{5}$$

Simplify the expression by subtracting \frac{36}{25} on both sides

$$-u^2 = -\frac{9}{5}-\frac{36}{25} = -\frac{81}{25}$$

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

$$u^2 = \frac{81}{25} u = \pm\sqrt{\frac{81}{25}} = \pm \frac{9}{5}$$

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

$$r =\frac{6}{5} - \frac{9}{5} = -0.600 s = \frac{6}{5} + \frac{9}{5} = 3$$

Evaluate

$$\left(j-3\right)\left(5j+3\right)$$