Solve 120x^2+67x-5

$$120 x ^ { 2 } + 67 x - 5$$

Factor

$$\left(15x-1\right)\left(8x+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 120x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=67 ab=120\left(-5\right)=-600$$

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 -600.

$$-1,600 -2,300 -3,200 -4,150 -5,120 -6,100 -8,75 -10,60 -12,50 -15,40 -20,30 -24,25$$

Calculate the sum for each pair.

$$-1+600=599 -2+300=298 -3+200=197 -4+150=146 -5+120=115 -6+100=94 -8+75=67 -10+60=50 -12+50=38 -15+40=25 -20+30=10 -24+25=1$$

The solution is the pair that gives sum 67.

$$a=-8 b=75$$

Rewrite 120x^{2}+67x-5 as \left(120x^{2}-8x\right)+\left(75x-5\right).

$$\left(120x^{2}-8x\right)+\left(75x-5\right)$$

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

$$8x\left(15x-1\right)+5\left(15x-1\right)$$

Factor out common term 15x-1 by using distributive property.

$$\left(15x-1\right)\left(8x+5\right)$$

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.

$$120x^{2}+67x-5=0$$

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.

$$x=\frac{-67±\sqrt{67^{2}-4\times 120\left(-5\right)}}{2\times 120}$$

Square 67.

$$x=\frac{-67±\sqrt{4489-4\times 120\left(-5\right)}}{2\times 120}$$

Multiply -4 times 120.

$$x=\frac{-67±\sqrt{4489-480\left(-5\right)}}{2\times 120}$$

Multiply -480 times -5.

$$x=\frac{-67±\sqrt{4489+2400}}{2\times 120}$$

Add 4489 to 2400.

$$x=\frac{-67±\sqrt{6889}}{2\times 120}$$

Take the square root of 6889.

$$x=\frac{-67±83}{2\times 120}$$

Multiply 2 times 120.

$$x=\frac{-67±83}{240}$$

Now solve the equation x=\frac{-67±83}{240} when ± is plus. Add -67 to 83.

$$x=\frac{16}{240}$$

Reduce the fraction \frac{16}{240} to lowest terms by extracting and canceling out 16.

$$x=\frac{1}{15}$$

Now solve the equation x=\frac{-67±83}{240} when ± is minus. Subtract 83 from -67.

$$x=-\frac{150}{240}$$

Reduce the fraction \frac{-150}{240} to lowest terms by extracting and canceling out 30.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1}{15} for x_{1} and -\frac{5}{8} for x_{2}.

$$120x^{2}+67x-5=120\left(x-\frac{1}{15}\right)\left(x-\left(-\frac{5}{8}\right)\right)$$

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

$$120x^{2}+67x-5=120\left(x-\frac{1}{15}\right)\left(x+\frac{5}{8}\right)$$

Subtract \frac{1}{15} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$120x^{2}+67x-5=120\times \frac{15x-1}{15}\left(x+\frac{5}{8}\right)$$

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

$$120x^{2}+67x-5=120\times \frac{15x-1}{15}\times \frac{8x+5}{8}$$

Multiply \frac{15x-1}{15} times \frac{8x+5}{8} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$120x^{2}+67x-5=120\times \frac{\left(15x-1\right)\left(8x+5\right)}{15\times 8}$$

Multiply 15 times 8.

$$120x^{2}+67x-5=120\times \frac{\left(15x-1\right)\left(8x+5\right)}{120}$$

Cancel out 120, the greatest common factor in 120 and 120.

$$120x^{2}+67x-5=\left(15x-1\right)\left(8x+5\right)$$

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 120

$$x ^ 2 +\frac{67}{120}x -\frac{1}{24} = 0$$

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

$$r + s = -\frac{67}{120} rs = -\frac{1}{24}$$

Two numbers r and s sum up to -\frac{67}{120} exactly when the average of the two numbers is \frac{1}{2}*-\frac{67}{120} = -\frac{67}{240}. 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>

$$r = -\frac{67}{240} - u s = -\frac{67}{240} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{24}

$$(-\frac{67}{240} - u) (-\frac{67}{240} + u) = -\frac{1}{24}$$

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

$$\frac{4489}{57600} - u^2 = -\frac{1}{24}$$

Simplify the expression by subtracting \frac{4489}{57600} on both sides

$$-u^2 = -\frac{1}{24}-\frac{4489}{57600} = -\frac{6889}{57600}$$

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

$$u^2 = \frac{6889}{57600} u = \pm\sqrt{\frac{6889}{57600}} = \pm \frac{83}{240}$$

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

$$r =-\frac{67}{240} - \frac{83}{240} = -0.625 s = -\frac{67}{240} + \frac{83}{240} = 0.067$$

Evaluate

$$\left(15x-1\right)\left(8x+5\right)$$