Solve 6x^2+29x-5

$$6 x ^ { 2 } + 29 x - 5$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 6x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=29 ab=6\left(-5\right)=-30$$

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 -30.

$$-1,30 -2,15 -3,10 -5,6$$

Calculate the sum for each pair.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

The solution is the pair that gives sum 29.

$$a=-1 b=30$$

Rewrite 6x^{2}+29x-5 as \left(6x^{2}-x\right)+\left(30x-5\right).

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

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

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

Factor out common term 6x-1 by using distributive property.

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

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.

$$6x^{2}+29x-5=0$$

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.

$$x=\frac{-29±\sqrt{29^{2}-4\times 6\left(-5\right)}}{2\times 6}$$

Square 29.

$$x=\frac{-29±\sqrt{841-4\times 6\left(-5\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-29±\sqrt{841-24\left(-5\right)}}{2\times 6}$$

Multiply -24 times -5.

$$x=\frac{-29±\sqrt{841+120}}{2\times 6}$$

Add 841 to 120.

$$x=\frac{-29±\sqrt{961}}{2\times 6}$$

Take the square root of 961.

$$x=\frac{-29±31}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{-29±31}{12}$$

Now solve the equation x=\frac{-29±31}{12} when ± is plus. Add -29 to 31.

$$x=\frac{2}{12}$$

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

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

Now solve the equation x=\frac{-29±31}{12} when ± is minus. Subtract 31 from -29.

$$x=-\frac{60}{12}$$

Divide -60 by 12.

$$x=-5$$

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

$$6x^{2}+29x-5=6\left(x-\frac{1}{6}\right)\left(x-\left(-5\right)\right)$$

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

$$6x^{2}+29x-5=6\left(x-\frac{1}{6}\right)\left(x+5\right)$$

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

$$6x^{2}+29x-5=6\times \frac{6x-1}{6}\left(x+5\right)$$

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

$$6x^{2}+29x-5=\left(6x-1\right)\left(x+5\right)$$

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 6

$$x ^ 2 +\frac{29}{6}x -\frac{5}{6} = 0$$

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

$$r + s = -\frac{29}{6} rs = -\frac{5}{6}$$

Two numbers r and s sum up to -\frac{29}{6} exactly when the average of the two numbers is \frac{1}{2}*-\frac{29}{6} = -\frac{29}{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>

$$r = -\frac{29}{12} - u s = -\frac{29}{12} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{6}

$$(-\frac{29}{12} - u) (-\frac{29}{12} + u) = -\frac{5}{6}$$

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

$$\frac{841}{144} - u^2 = -\frac{5}{6}$$

Simplify the expression by subtracting \frac{841}{144} on both sides

$$-u^2 = -\frac{5}{6}-\frac{841}{144} = -\frac{961}{144}$$

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

$$u^2 = \frac{961}{144} u = \pm\sqrt{\frac{961}{144}} = \pm \frac{31}{12}$$

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

$$r =-\frac{29}{12} - \frac{31}{12} = -5 s = -\frac{29}{12} + \frac{31}{12} = 0.167$$

Evaluate

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