Solve 6r^2+29r-42

$$6 r ^ { 2 } + 29 r - 42$$

Factor

$$\left(6r-7\right)\left(r+6\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 6r^{2}+ar+br-42. To find a and b, set up a system to be solved.

$$a+b=29 ab=6\left(-42\right)=-252$$

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 -252.

$$-1,252 -2,126 -3,84 -4,63 -6,42 -7,36 -9,28 -12,21 -14,18$$

Calculate the sum for each pair.

$$-1+252=251 -2+126=124 -3+84=81 -4+63=59 -6+42=36 -7+36=29 -9+28=19 -12+21=9 -14+18=4$$

The solution is the pair that gives sum 29.

$$a=-7 b=36$$

Rewrite 6r^{2}+29r-42 as \left(6r^{2}-7r\right)+\left(36r-42\right).

$$\left(6r^{2}-7r\right)+\left(36r-42\right)$$

Factor out r in the first and 6 in the second group.

$$r\left(6r-7\right)+6\left(6r-7\right)$$

Factor out common term 6r-7 by using distributive property.

$$\left(6r-7\right)\left(r+6\right)$$

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.

$$6r^{2}+29r-42=0$$

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.

$$r=\frac{-29±\sqrt{29^{2}-4\times 6\left(-42\right)}}{2\times 6}$$

Square 29.

$$r=\frac{-29±\sqrt{841-4\times 6\left(-42\right)}}{2\times 6}$$

Multiply -4 times 6.

$$r=\frac{-29±\sqrt{841-24\left(-42\right)}}{2\times 6}$$

Multiply -24 times -42.

$$r=\frac{-29±\sqrt{841+1008}}{2\times 6}$$

Add 841 to 1008.

$$r=\frac{-29±\sqrt{1849}}{2\times 6}$$

Take the square root of 1849.

$$r=\frac{-29±43}{2\times 6}$$

Multiply 2 times 6.

$$r=\frac{-29±43}{12}$$

Now solve the equation r=\frac{-29±43}{12} when ± is plus. Add -29 to 43.

$$r=\frac{14}{12}$$

Reduce the fraction \frac{14}{12} to lowest terms by extracting and canceling out 2.

$$r=\frac{7}{6}$$

Now solve the equation r=\frac{-29±43}{12} when ± is minus. Subtract 43 from -29.

$$r=-\frac{72}{12}$$

Divide -72 by 12.

$$r=-6$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{7}{6} for x_{1} and -6 for x_{2}.

$$6r^{2}+29r-42=6\left(r-\frac{7}{6}\right)\left(r-\left(-6\right)\right)$$

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

$$6r^{2}+29r-42=6\left(r-\frac{7}{6}\right)\left(r+6\right)$$

Subtract \frac{7}{6} from r by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$6r^{2}+29r-42=6\times \frac{6r-7}{6}\left(r+6\right)$$

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

$$6r^{2}+29r-42=\left(6r-7\right)\left(r+6\right)$$

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

$$x ^ 2 +\frac{29}{6}x -7 = 0$$

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

$$r + s = -\frac{29}{6} rs = -7$$

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 = -7

$$(-\frac{29}{12} - u) (-\frac{29}{12} + u) = -7$$

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

$$\frac{841}{144} - u^2 = -7$$

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

$$-u^2 = -7-\frac{841}{144} = -\frac{1849}{144}$$

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

$$u^2 = \frac{1849}{144} u = \pm\sqrt{\frac{1849}{144}} = \pm \frac{43}{12}$$

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

$$r =-\frac{29}{12} - \frac{43}{12} = -6 s = -\frac{29}{12} + \frac{43}{12} = 1.167$$

Evaluate

$$\left(6r-7\right)\left(r+6\right)$$