Solve 30x^2+7x-15

$$30 x ^ { 2 } + 7 x - 15$$

Factor

$$\left(5x-3\right)\left(6x+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 30x^{2}+ax+bx-15. To find a and b, set up a system to be solved.

$$a+b=7 ab=30\left(-15\right)=-450$$

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 -450.

$$-1,450 -2,225 -3,150 -5,90 -6,75 -9,50 -10,45 -15,30 -18,25$$

Calculate the sum for each pair.

$$-1+450=449 -2+225=223 -3+150=147 -5+90=85 -6+75=69 -9+50=41 -10+45=35 -15+30=15 -18+25=7$$

The solution is the pair that gives sum 7.

$$a=-18 b=25$$

Rewrite 30x^{2}+7x-15 as \left(30x^{2}-18x\right)+\left(25x-15\right).

$$\left(30x^{2}-18x\right)+\left(25x-15\right)$$

Factor out 6x in the first and 5 in the second group.

$$6x\left(5x-3\right)+5\left(5x-3\right)$$

Factor out common term 5x-3 by using distributive property.

$$\left(5x-3\right)\left(6x+5\right)$$

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.

$$30x^{2}+7x-15=0$$

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.

$$x=\frac{-7±\sqrt{7^{2}-4\times 30\left(-15\right)}}{2\times 30}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\times 30\left(-15\right)}}{2\times 30}$$

Multiply -4 times 30.

$$x=\frac{-7±\sqrt{49-120\left(-15\right)}}{2\times 30}$$

Multiply -120 times -15.

$$x=\frac{-7±\sqrt{49+1800}}{2\times 30}$$

Add 49 to 1800.

$$x=\frac{-7±\sqrt{1849}}{2\times 30}$$

Take the square root of 1849.

$$x=\frac{-7±43}{2\times 30}$$

Multiply 2 times 30.

$$x=\frac{-7±43}{60}$$

Now solve the equation x=\frac{-7±43}{60} when ± is plus. Add -7 to 43.

$$x=\frac{36}{60}$$

Reduce the fraction \frac{36}{60} to lowest terms by extracting and canceling out 12.

$$x=\frac{3}{5}$$

Now solve the equation x=\frac{-7±43}{60} when ± is minus. Subtract 43 from -7.

$$x=-\frac{50}{60}$$

Reduce the fraction \frac{-50}{60} to lowest terms by extracting and canceling out 10.

$$x=-\frac{5}{6}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3}{5} for x_{1} and -\frac{5}{6} for x_{2}.

$$30x^{2}+7x-15=30\left(x-\frac{3}{5}\right)\left(x-\left(-\frac{5}{6}\right)\right)$$

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

$$30x^{2}+7x-15=30\left(x-\frac{3}{5}\right)\left(x+\frac{5}{6}\right)$$

Subtract \frac{3}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$30x^{2}+7x-15=30\times \frac{5x-3}{5}\left(x+\frac{5}{6}\right)$$

Add \frac{5}{6} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$30x^{2}+7x-15=30\times \frac{5x-3}{5}\times \frac{6x+5}{6}$$

Multiply \frac{5x-3}{5} times \frac{6x+5}{6} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$30x^{2}+7x-15=30\times \frac{\left(5x-3\right)\left(6x+5\right)}{5\times 6}$$

Multiply 5 times 6.

$$30x^{2}+7x-15=30\times \frac{\left(5x-3\right)\left(6x+5\right)}{30}$$

Cancel out 30, the greatest common factor in 30 and 30.

$$30x^{2}+7x-15=\left(5x-3\right)\left(6x+5\right)$$

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 30

$$x ^ 2 +\frac{7}{30}x -\frac{1}{2} = 0$$

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

$$r + s = -\frac{7}{30} rs = -\frac{1}{2}$$

Two numbers r and s sum up to -\frac{7}{30} exactly when the average of the two numbers is \frac{1}{2}*-\frac{7}{30} = -\frac{7}{60}. 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>

$$r = -\frac{7}{60} - u s = -\frac{7}{60} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{2}

$$(-\frac{7}{60} - u) (-\frac{7}{60} + u) = -\frac{1}{2}$$

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

$$\frac{49}{3600} - u^2 = -\frac{1}{2}$$

Simplify the expression by subtracting \frac{49}{3600} on both sides

$$-u^2 = -\frac{1}{2}-\frac{49}{3600} = -\frac{1849}{3600}$$

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

$$u^2 = \frac{1849}{3600} u = \pm\sqrt{\frac{1849}{3600}} = \pm \frac{43}{60}$$

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

$$r =-\frac{7}{60} - \frac{43}{60} = -0.833 s = -\frac{7}{60} + \frac{43}{60} = 0.600$$

Evaluate

$$\left(5x-3\right)\left(6x+5\right)$$