Solve 6x^2-9x-15

$$6 x ^ { 2 } - 9 x - 15$$

Factor

$$3\left(2x-5\right)\left(x+1\right)$$

Solution Steps

Factor out 3.

$$3\left(2x^{2}-3x-5\right)$$

Consider 2x^{2}-3x-5. Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=-3 ab=2\left(-5\right)=-10$$

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 -10.

$$1,-10 2,-5$$

Calculate the sum for each pair.

$$1-10=-9 2-5=-3$$

The solution is the pair that gives sum -3.

$$a=-5 b=2$$

Rewrite 2x^{2}-3x-5 as \left(2x^{2}-5x\right)+\left(2x-5\right).

$$\left(2x^{2}-5x\right)+\left(2x-5\right)$$

Factor out x in 2x^{2}-5x.

$$x\left(2x-5\right)+2x-5$$

Factor out common term 2x-5 by using distributive property.

$$\left(2x-5\right)\left(x+1\right)$$

Rewrite the complete factored expression.

$$3\left(2x-5\right)\left(x+1\right)$$

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.

$$6x^{2}-9x-15=0$$

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.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 6\left(-15\right)}}{2\times 6}$$

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 6\left(-15\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-9\right)±\sqrt{81-24\left(-15\right)}}{2\times 6}$$

Multiply -24 times -15.

$$x=\frac{-\left(-9\right)±\sqrt{81+360}}{2\times 6}$$

Add 81 to 360.

$$x=\frac{-\left(-9\right)±\sqrt{441}}{2\times 6}$$

Take the square root of 441.

$$x=\frac{-\left(-9\right)±21}{2\times 6}$$

The opposite of -9 is 9.

$$x=\frac{9±21}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{9±21}{12}$$

Now solve the equation x=\frac{9±21}{12} when ± is plus. Add 9 to 21.

$$x=\frac{30}{12}$$

Reduce the fraction \frac{30}{12} to lowest terms by extracting and canceling out 6.

$$x=\frac{5}{2}$$

Now solve the equation x=\frac{9±21}{12} when ± is minus. Subtract 21 from 9.

$$x=-\frac{12}{12}$$

Divide -12 by 12.

$$x=-1$$

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

$$6x^{2}-9x-15=6\left(x-\frac{5}{2}\right)\left(x-\left(-1\right)\right)$$

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

$$6x^{2}-9x-15=6\left(x-\frac{5}{2}\right)\left(x+1\right)$$

Subtract \frac{5}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$6x^{2}-9x-15=6\times \frac{2x-5}{2}\left(x+1\right)$$

Cancel out 2, the greatest common factor in 6 and 2.

$$6x^{2}-9x-15=3\left(2x-5\right)\left(x+1\right)$$

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 6

$$x ^ 2 -\frac{3}{2}x -\frac{5}{2} = 0$$

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

$$r + s = \frac{3}{2} rs = -\frac{5}{2}$$

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

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

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{2}

$$(\frac{3}{4} - u) (\frac{3}{4} + u) = -\frac{5}{2}$$

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

$$\frac{9}{16} - u^2 = -\frac{5}{2}$$

Simplify the expression by subtracting \frac{9}{16} on both sides

$$-u^2 = -\frac{5}{2}-\frac{9}{16} = -\frac{49}{16}$$

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

$$u^2 = \frac{49}{16} u = \pm\sqrt{\frac{49}{16}} = \pm \frac{7}{4}$$

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

$$r =\frac{3}{4} - \frac{7}{4} = -1 s = \frac{3}{4} + \frac{7}{4} = 2.500$$

Evaluate

$$3\left(2x-5\right)\left(x+1\right)$$