Solve 15x^2-21x-72

$$15 x ^ { 2 } - 21 x - 72$$

Factor

$$3\left(x-3\right)\left(5x+8\right)$$

Solution Steps

Factor out 3.

$$3\left(5x^{2}-7x-24\right)$$

Consider 5x^{2}-7x-24. Factor the expression by grouping. First, the expression needs to be rewritten as 5x^{2}+ax+bx-24. To find a and b, set up a system to be solved.

$$a+b=-7 ab=5\left(-24\right)=-120$$

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 -120.

$$1,-120 2,-60 3,-40 4,-30 5,-24 6,-20 8,-15 10,-12$$

Calculate the sum for each pair.

$$1-120=-119 2-60=-58 3-40=-37 4-30=-26 5-24=-19 6-20=-14 8-15=-7 10-12=-2$$

The solution is the pair that gives sum -7.

$$a=-15 b=8$$

Rewrite 5x^{2}-7x-24 as \left(5x^{2}-15x\right)+\left(8x-24\right).

$$\left(5x^{2}-15x\right)+\left(8x-24\right)$$

Factor out 5x in the first and 8 in the second group.

$$5x\left(x-3\right)+8\left(x-3\right)$$

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

$$\left(x-3\right)\left(5x+8\right)$$

Rewrite the complete factored expression.

$$3\left(x-3\right)\left(5x+8\right)$$

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.

$$15x^{2}-21x-72=0$$

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.

$$x=\frac{-\left(-21\right)±\sqrt{\left(-21\right)^{2}-4\times 15\left(-72\right)}}{2\times 15}$$

Square -21.

$$x=\frac{-\left(-21\right)±\sqrt{441-4\times 15\left(-72\right)}}{2\times 15}$$

Multiply -4 times 15.

$$x=\frac{-\left(-21\right)±\sqrt{441-60\left(-72\right)}}{2\times 15}$$

Multiply -60 times -72.

$$x=\frac{-\left(-21\right)±\sqrt{441+4320}}{2\times 15}$$

Add 441 to 4320.

$$x=\frac{-\left(-21\right)±\sqrt{4761}}{2\times 15}$$

Take the square root of 4761.

$$x=\frac{-\left(-21\right)±69}{2\times 15}$$

The opposite of -21 is 21.

$$x=\frac{21±69}{2\times 15}$$

Multiply 2 times 15.

$$x=\frac{21±69}{30}$$

Now solve the equation x=\frac{21±69}{30} when ± is plus. Add 21 to 69.

$$x=\frac{90}{30}$$

Divide 90 by 30.

$$x=3$$

Now solve the equation x=\frac{21±69}{30} when ± is minus. Subtract 69 from 21.

$$x=-\frac{48}{30}$$

Reduce the fraction \frac{-48}{30} to lowest terms by extracting and canceling out 6.

$$x=-\frac{8}{5}$$

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{8}{5} for x_{2}.

$$15x^{2}-21x-72=15\left(x-3\right)\left(x-\left(-\frac{8}{5}\right)\right)$$

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

$$15x^{2}-21x-72=15\left(x-3\right)\left(x+\frac{8}{5}\right)$$

Add \frac{8}{5} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$15x^{2}-21x-72=15\left(x-3\right)\times \frac{5x+8}{5}$$

Cancel out 5, the greatest common factor in 15 and 5.

$$15x^{2}-21x-72=3\left(x-3\right)\left(5x+8\right)$$

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 15

$$x ^ 2 -\frac{7}{5}x -\frac{24}{5} = 0$$

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

$$r + s = \frac{7}{5} rs = -\frac{24}{5}$$

Two numbers r and s sum up to \frac{7}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{7}{5} = \frac{7}{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>

$$r = \frac{7}{10} - u s = \frac{7}{10} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{24}{5}

$$(\frac{7}{10} - u) (\frac{7}{10} + u) = -\frac{24}{5}$$

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

$$\frac{49}{100} - u^2 = -\frac{24}{5}$$

Simplify the expression by subtracting \frac{49}{100} on both sides

$$-u^2 = -\frac{24}{5}-\frac{49}{100} = -\frac{529}{100}$$

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.

$$r =\frac{7}{10} - \frac{23}{10} = -1.600 s = \frac{7}{10} + \frac{23}{10} = 3$$

Evaluate

$$3\left(x-3\right)\left(5x+8\right)$$