Solve 30x^2-7x-2

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

Factor

$$\left(5x-2\right)\left(6x+1\right)$$

Solution Steps

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

$$a+b=-7 ab=30\left(-2\right)=-60$$

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 -60.

$$1,-60 2,-30 3,-20 4,-15 5,-12 6,-10$$

Calculate the sum for each pair.

$$1-60=-59 2-30=-28 3-20=-17 4-15=-11 5-12=-7 6-10=-4$$

The solution is the pair that gives sum -7.

$$a=-12 b=5$$

Rewrite 30x^{2}-7x-2 as \left(30x^{2}-12x\right)+\left(5x-2\right).

$$\left(30x^{2}-12x\right)+\left(5x-2\right)$$

Factor out 6x in 30x^{2}-12x.

$$6x\left(5x-2\right)+5x-2$$

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

$$\left(5x-2\right)\left(6x+1\right)$$

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.

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

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.

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

Square -7.

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

Multiply -4 times 30.

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

Multiply -120 times -2.

$$x=\frac{-\left(-7\right)±\sqrt{49+240}}{2\times 30}$$

Add 49 to 240.

$$x=\frac{-\left(-7\right)±\sqrt{289}}{2\times 30}$$

Take the square root of 289.

$$x=\frac{-\left(-7\right)±17}{2\times 30}$$

The opposite of -7 is 7.

$$x=\frac{7±17}{2\times 30}$$

Multiply 2 times 30.

$$x=\frac{7±17}{60}$$

Now solve the equation x=\frac{7±17}{60} when ± is plus. Add 7 to 17.

$$x=\frac{24}{60}$$

Reduce the fraction \frac{24}{60} to lowest terms by extracting and canceling out 12.

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

Now solve the equation x=\frac{7±17}{60} when ± is minus. Subtract 17 from 7.

$$x=-\frac{10}{60}$$

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

$$x=-\frac{1}{6}$$

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 -\frac{1}{6} for x_{2}.

$$30x^{2}-7x-2=30\left(x-\frac{2}{5}\right)\left(x-\left(-\frac{1}{6}\right)\right)$$

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

$$30x^{2}-7x-2=30\left(x-\frac{2}{5}\right)\left(x+\frac{1}{6}\right)$$

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

$$30x^{2}-7x-2=30\times \frac{5x-2}{5}\left(x+\frac{1}{6}\right)$$

Add \frac{1}{6} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$30x^{2}-7x-2=30\times \frac{5x-2}{5}\times \frac{6x+1}{6}$$

Multiply \frac{5x-2}{5} times \frac{6x+1}{6} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$30x^{2}-7x-2=30\times \frac{\left(5x-2\right)\left(6x+1\right)}{5\times 6}$$

Multiply 5 times 6.

$$30x^{2}-7x-2=30\times \frac{\left(5x-2\right)\left(6x+1\right)}{30}$$

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

$$30x^{2}-7x-2=\left(5x-2\right)\left(6x+1\right)$$

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 30

$$x ^ 2 -\frac{7}{30}x -\frac{1}{15} = 0$$

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

$$r + s = \frac{7}{30} rs = -\frac{1}{15}$$

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}{15}

$$(\frac{7}{60} - u) (\frac{7}{60} + u) = -\frac{1}{15}$$

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

$$\frac{49}{3600} - u^2 = -\frac{1}{15}$$

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

$$-u^2 = -\frac{1}{15}-\frac{49}{3600} = -\frac{289}{3600}$$

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

$$u^2 = \frac{289}{3600} u = \pm\sqrt{\frac{289}{3600}} = \pm \frac{17}{60}$$

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

$$r =\frac{7}{60} - \frac{17}{60} = -0.167 s = \frac{7}{60} + \frac{17}{60} = 0.400$$

Evaluate

$$\left(5x-2\right)\left(6x+1\right)$$