Solve 6n^2-17n-3

$$6 n ^ { 2 } - 17 n - 3$$

Factor

$$\left(n-3\right)\left(6n+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 6n^{2}+an+bn-3. To find a and b, set up a system to be solved.

$$a+b=-17 ab=6\left(-3\right)=-18$$

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 -18.

$$1,-18 2,-9 3,-6$$

Calculate the sum for each pair.

$$1-18=-17 2-9=-7 3-6=-3$$

The solution is the pair that gives sum -17.

$$a=-18 b=1$$

Rewrite 6n^{2}-17n-3 as \left(6n^{2}-18n\right)+\left(n-3\right).

$$\left(6n^{2}-18n\right)+\left(n-3\right)$$

Factor out 6n in 6n^{2}-18n.

$$6n\left(n-3\right)+n-3$$

Factor out common term n-3 by using distributive property.

$$\left(n-3\right)\left(6n+1\right)$$

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.

$$6n^{2}-17n-3=0$$

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.

$$n=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 6\left(-3\right)}}{2\times 6}$$

Square -17.

$$n=\frac{-\left(-17\right)±\sqrt{289-4\times 6\left(-3\right)}}{2\times 6}$$

Multiply -4 times 6.

$$n=\frac{-\left(-17\right)±\sqrt{289-24\left(-3\right)}}{2\times 6}$$

Multiply -24 times -3.

$$n=\frac{-\left(-17\right)±\sqrt{289+72}}{2\times 6}$$

Add 289 to 72.

$$n=\frac{-\left(-17\right)±\sqrt{361}}{2\times 6}$$

Take the square root of 361.

$$n=\frac{-\left(-17\right)±19}{2\times 6}$$

The opposite of -17 is 17.

$$n=\frac{17±19}{2\times 6}$$

Multiply 2 times 6.

$$n=\frac{17±19}{12}$$

Now solve the equation n=\frac{17±19}{12} when ± is plus. Add 17 to 19.

$$n=\frac{36}{12}$$

Divide 36 by 12.

$$n=3$$

Now solve the equation n=\frac{17±19}{12} when ± is minus. Subtract 19 from 17.

$$n=-\frac{2}{12}$$

Reduce the fraction \frac{-2}{12} to lowest terms by extracting and canceling out 2.

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

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

$$6n^{2}-17n-3=6\left(n-3\right)\left(n-\left(-\frac{1}{6}\right)\right)$$

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

$$6n^{2}-17n-3=6\left(n-3\right)\left(n+\frac{1}{6}\right)$$

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

$$6n^{2}-17n-3=6\left(n-3\right)\times \frac{6n+1}{6}$$

Cancel out 6, the greatest common factor in 6 and 6.

$$6n^{2}-17n-3=\left(n-3\right)\left(6n+1\right)$$

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 6

$$x ^ 2 -\frac{17}{6}x -\frac{1}{2} = 0$$

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

$$r + s = \frac{17}{6} rs = -\frac{1}{2}$$

Two numbers r and s sum up to \frac{17}{6} exactly when the average of the two numbers is \frac{1}{2}*\frac{17}{6} = \frac{17}{12}. 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>

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

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

$$(\frac{17}{12} - u) (\frac{17}{12} + u) = -\frac{1}{2}$$

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

$$\frac{289}{144} - u^2 = -\frac{1}{2}$$

Simplify the expression by subtracting \frac{289}{144} on both sides

$$-u^2 = -\frac{1}{2}-\frac{289}{144} = -\frac{361}{144}$$

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

$$u^2 = \frac{361}{144} u = \pm\sqrt{\frac{361}{144}} = \pm \frac{19}{12}$$

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

$$r =\frac{17}{12} - \frac{19}{12} = -0.167 s = \frac{17}{12} + \frac{19}{12} = 3$$

Evaluate

$$\left(n-3\right)\left(6n+1\right)$$