Solve 12x^2-17x-5

$$12 x ^ { 2 } - 17 x - 5$$

Factor

$$\left(3x-5\right)\left(4x+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 12x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=-17 ab=12\left(-5\right)=-60$$

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 -17.

$$a=-20 b=3$$

Rewrite 12x^{2}-17x-5 as \left(12x^{2}-20x\right)+\left(3x-5\right).

$$\left(12x^{2}-20x\right)+\left(3x-5\right)$$

Factor out 4x in 12x^{2}-20x.

$$4x\left(3x-5\right)+3x-5$$

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

$$\left(3x-5\right)\left(4x+1\right)$$

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.

$$12x^{2}-17x-5=0$$

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.

$$x=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 12\left(-5\right)}}{2\times 12}$$

Square -17.

$$x=\frac{-\left(-17\right)±\sqrt{289-4\times 12\left(-5\right)}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-\left(-17\right)±\sqrt{289-48\left(-5\right)}}{2\times 12}$$

Multiply -48 times -5.

$$x=\frac{-\left(-17\right)±\sqrt{289+240}}{2\times 12}$$

Add 289 to 240.

$$x=\frac{-\left(-17\right)±\sqrt{529}}{2\times 12}$$

Take the square root of 529.

$$x=\frac{-\left(-17\right)±23}{2\times 12}$$

The opposite of -17 is 17.

$$x=\frac{17±23}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{17±23}{24}$$

Now solve the equation x=\frac{17±23}{24} when ± is plus. Add 17 to 23.

$$x=\frac{40}{24}$$

Reduce the fraction \frac{40}{24} to lowest terms by extracting and canceling out 8.

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

Now solve the equation x=\frac{17±23}{24} when ± is minus. Subtract 23 from 17.

$$x=-\frac{6}{24}$$

Reduce the fraction \frac{-6}{24} to lowest terms by extracting and canceling out 6.

$$x=-\frac{1}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{3} for x_{1} and -\frac{1}{4} for x_{2}.

$$12x^{2}-17x-5=12\left(x-\frac{5}{3}\right)\left(x-\left(-\frac{1}{4}\right)\right)$$

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

$$12x^{2}-17x-5=12\left(x-\frac{5}{3}\right)\left(x+\frac{1}{4}\right)$$

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

$$12x^{2}-17x-5=12\times \frac{3x-5}{3}\left(x+\frac{1}{4}\right)$$

Add \frac{1}{4} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$12x^{2}-17x-5=12\times \frac{3x-5}{3}\times \frac{4x+1}{4}$$

Multiply \frac{3x-5}{3} times \frac{4x+1}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$12x^{2}-17x-5=12\times \frac{\left(3x-5\right)\left(4x+1\right)}{3\times 4}$$

Multiply 3 times 4.

$$12x^{2}-17x-5=12\times \frac{\left(3x-5\right)\left(4x+1\right)}{12}$$

Cancel out 12, the greatest common factor in 12 and 12.

$$12x^{2}-17x-5=\left(3x-5\right)\left(4x+1\right)$$

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 12

$$x ^ 2 -\frac{17}{12}x -\frac{5}{12} = 0$$

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

$$r + s = \frac{17}{12} rs = -\frac{5}{12}$$

Two numbers r and s sum up to \frac{17}{12} exactly when the average of the two numbers is \frac{1}{2}*\frac{17}{12} = \frac{17}{24}. 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>

$$r = \frac{17}{24} - u s = \frac{17}{24} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{12}

$$(\frac{17}{24} - u) (\frac{17}{24} + u) = -\frac{5}{12}$$

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

$$\frac{289}{576} - u^2 = -\frac{5}{12}$$

Simplify the expression by subtracting \frac{289}{576} on both sides

$$-u^2 = -\frac{5}{12}-\frac{289}{576} = -\frac{529}{576}$$

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

$$u^2 = \frac{529}{576} u = \pm\sqrt{\frac{529}{576}} = \pm \frac{23}{24}$$

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

$$r =\frac{17}{24} - \frac{23}{24} = -0.250 s = \frac{17}{24} + \frac{23}{24} = 1.667$$

Evaluate

$$\left(3x-5\right)\left(4x+1\right)$$