Solve 15x^2+69x-30

$$15 x ^ { 2 } + 69 x - 30$$

Factor

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

Solution Steps

Factor out 3.

$$3\left(5x^{2}+23x-10\right)$$

Consider 5x^{2}+23x-10. Factor the expression by grouping. First, the expression needs to be rewritten as 5x^{2}+ax+bx-10. To find a and b, set up a system to be solved.

$$a+b=23 ab=5\left(-10\right)=-50$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -50.

$$-1,50 -2,25 -5,10$$

Calculate the sum for each pair.

$$-1+50=49 -2+25=23 -5+10=5$$

The solution is the pair that gives sum 23.

$$a=-2 b=25$$

Rewrite 5x^{2}+23x-10 as \left(5x^{2}-2x\right)+\left(25x-10\right).

$$\left(5x^{2}-2x\right)+\left(25x-10\right)$$

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

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

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

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

Rewrite the complete factored expression.

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

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.

$$15x^{2}+69x-30=0$$

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.

$$x=\frac{-69±\sqrt{69^{2}-4\times 15\left(-30\right)}}{2\times 15}$$

Square 69.

$$x=\frac{-69±\sqrt{4761-4\times 15\left(-30\right)}}{2\times 15}$$

Multiply -4 times 15.

$$x=\frac{-69±\sqrt{4761-60\left(-30\right)}}{2\times 15}$$

Multiply -60 times -30.

$$x=\frac{-69±\sqrt{4761+1800}}{2\times 15}$$

Add 4761 to 1800.

$$x=\frac{-69±\sqrt{6561}}{2\times 15}$$

Take the square root of 6561.

$$x=\frac{-69±81}{2\times 15}$$

Multiply 2 times 15.

$$x=\frac{-69±81}{30}$$

Now solve the equation x=\frac{-69±81}{30} when ± is plus. Add -69 to 81.

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

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

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

Now solve the equation x=\frac{-69±81}{30} when ± is minus. Subtract 81 from -69.

$$x=-\frac{150}{30}$$

Divide -150 by 30.

$$x=-5$$

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

$$15x^{2}+69x-30=15\left(x-\frac{2}{5}\right)\left(x-\left(-5\right)\right)$$

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

$$15x^{2}+69x-30=15\left(x-\frac{2}{5}\right)\left(x+5\right)$$

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

$$15x^{2}+69x-30=15\times \frac{5x-2}{5}\left(x+5\right)$$

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

$$15x^{2}+69x-30=3\left(5x-2\right)\left(x+5\right)$$

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 15

$$x ^ 2 +\frac{23}{5}x -2 = 0$$

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

$$r + s = -\frac{23}{5} rs = -2$$

Two numbers r and s sum up to -\frac{23}{5} exactly when the average of the two numbers is \frac{1}{2}*-\frac{23}{5} = -\frac{23}{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>

$$r = -\frac{23}{10} - u s = -\frac{23}{10} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -2

$$(-\frac{23}{10} - u) (-\frac{23}{10} + u) = -2$$

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

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

Simplify the expression by subtracting \frac{529}{100} on both sides

$$-u^2 = -2-\frac{529}{100} = -\frac{729}{100}$$

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

$$u^2 = \frac{729}{100} u = \pm\sqrt{\frac{729}{100}} = \pm \frac{27}{10}$$

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

$$r =-\frac{23}{10} - \frac{27}{10} = -5 s = -\frac{23}{10} + \frac{27}{10} = 0.400$$

Evaluate

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