Solve 6r^2-35r-49

$$6 r ^ { 2 } - 35 r - 49$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 6r^{2}+ar+br-49. To find a and b, set up a system to be solved.

$$a+b=-35 ab=6\left(-49\right)=-294$$

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 -294.

$$1,-294 2,-147 3,-98 6,-49 7,-42 14,-21$$

Calculate the sum for each pair.

$$1-294=-293 2-147=-145 3-98=-95 6-49=-43 7-42=-35 14-21=-7$$

The solution is the pair that gives sum -35.

$$a=-42 b=7$$

Rewrite 6r^{2}-35r-49 as \left(6r^{2}-42r\right)+\left(7r-49\right).

$$\left(6r^{2}-42r\right)+\left(7r-49\right)$$

Factor out 6r in the first and 7 in the second group.

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

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

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

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.

$$6r^{2}-35r-49=0$$

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.

$$r=\frac{-\left(-35\right)±\sqrt{\left(-35\right)^{2}-4\times 6\left(-49\right)}}{2\times 6}$$

Square -35.

$$r=\frac{-\left(-35\right)±\sqrt{1225-4\times 6\left(-49\right)}}{2\times 6}$$

Multiply -4 times 6.

$$r=\frac{-\left(-35\right)±\sqrt{1225-24\left(-49\right)}}{2\times 6}$$

Multiply -24 times -49.

$$r=\frac{-\left(-35\right)±\sqrt{1225+1176}}{2\times 6}$$

Add 1225 to 1176.

$$r=\frac{-\left(-35\right)±\sqrt{2401}}{2\times 6}$$

Take the square root of 2401.

$$r=\frac{-\left(-35\right)±49}{2\times 6}$$

The opposite of -35 is 35.

$$r=\frac{35±49}{2\times 6}$$

Multiply 2 times 6.

$$r=\frac{35±49}{12}$$

Now solve the equation r=\frac{35±49}{12} when ± is plus. Add 35 to 49.

$$r=\frac{84}{12}$$

Divide 84 by 12.

$$r=7$$

Now solve the equation r=\frac{35±49}{12} when ± is minus. Subtract 49 from 35.

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

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

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

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

$$6r^{2}-35r-49=6\left(r-7\right)\left(r-\left(-\frac{7}{6}\right)\right)$$

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

$$6r^{2}-35r-49=6\left(r-7\right)\left(r+\frac{7}{6}\right)$$

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

$$6r^{2}-35r-49=6\left(r-7\right)\times \frac{6r+7}{6}$$

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

$$6r^{2}-35r-49=\left(r-7\right)\left(6r+7\right)$$

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

$$x ^ 2 -\frac{35}{6}x -\frac{49}{6} = 0$$

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

$$r + s = \frac{35}{6} rs = -\frac{49}{6}$$

Two numbers r and s sum up to \frac{35}{6} exactly when the average of the two numbers is \frac{1}{2}*\frac{35}{6} = \frac{35}{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>

$$r = \frac{35}{12} - u s = \frac{35}{12} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{49}{6}

$$(\frac{35}{12} - u) (\frac{35}{12} + u) = -\frac{49}{6}$$

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

$$\frac{1225}{144} - u^2 = -\frac{49}{6}$$

Simplify the expression by subtracting \frac{1225}{144} on both sides

$$-u^2 = -\frac{49}{6}-\frac{1225}{144} = -\frac{2401}{144}$$

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

$$u^2 = \frac{2401}{144} u = \pm\sqrt{\frac{2401}{144}} = \pm \frac{49}{12}$$

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

$$r =\frac{35}{12} - \frac{49}{12} = -1.167 s = \frac{35}{12} + \frac{49}{12} = 7$$

Evaluate

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