Solve 6r^2+53r-70

$$6 r ^ { 2 } + 53 r - 70$$

Factor

$$\left(6r-7\right)\left(r+10\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 6r^{2}+ar+br-70. To find a and b, set up a system to be solved.

$$a+b=53 ab=6\left(-70\right)=-420$$

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 -420.

$$-1,420 -2,210 -3,140 -4,105 -5,84 -6,70 -7,60 -10,42 -12,35 -14,30 -15,28 -20,21$$

Calculate the sum for each pair.

$$-1+420=419 -2+210=208 -3+140=137 -4+105=101 -5+84=79 -6+70=64 -7+60=53 -10+42=32 -12+35=23 -14+30=16 -15+28=13 -20+21=1$$

The solution is the pair that gives sum 53.

$$a=-7 b=60$$

Rewrite 6r^{2}+53r-70 as \left(6r^{2}-7r\right)+\left(60r-70\right).

$$\left(6r^{2}-7r\right)+\left(60r-70\right)$$

Factor out r in the first and 10 in the second group.

$$r\left(6r-7\right)+10\left(6r-7\right)$$

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

$$\left(6r-7\right)\left(r+10\right)$$

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.

$$6r^{2}+53r-70=0$$

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.

$$r=\frac{-53±\sqrt{53^{2}-4\times 6\left(-70\right)}}{2\times 6}$$

Square 53.

$$r=\frac{-53±\sqrt{2809-4\times 6\left(-70\right)}}{2\times 6}$$

Multiply -4 times 6.

$$r=\frac{-53±\sqrt{2809-24\left(-70\right)}}{2\times 6}$$

Multiply -24 times -70.

$$r=\frac{-53±\sqrt{2809+1680}}{2\times 6}$$

Add 2809 to 1680.

$$r=\frac{-53±\sqrt{4489}}{2\times 6}$$

Take the square root of 4489.

$$r=\frac{-53±67}{2\times 6}$$

Multiply 2 times 6.

$$r=\frac{-53±67}{12}$$

Now solve the equation r=\frac{-53±67}{12} when ± is plus. Add -53 to 67.

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

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{-53±67}{12} when ± is minus. Subtract 67 from -53.

$$r=-\frac{120}{12}$$

Divide -120 by 12.

$$r=-10$$

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 -10 for x_{2}.

$$6r^{2}+53r-70=6\left(r-\frac{7}{6}\right)\left(r-\left(-10\right)\right)$$

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

$$6r^{2}+53r-70=6\left(r-\frac{7}{6}\right)\left(r+10\right)$$

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

$$6r^{2}+53r-70=6\times \frac{6r-7}{6}\left(r+10\right)$$

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

$$6r^{2}+53r-70=\left(6r-7\right)\left(r+10\right)$$

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

$$x ^ 2 +\frac{53}{6}x -\frac{35}{3} = 0$$

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

$$r + s = -\frac{53}{6} rs = -\frac{35}{3}$$

Two numbers r and s sum up to -\frac{53}{6} exactly when the average of the two numbers is \frac{1}{2}*-\frac{53}{6} = -\frac{53}{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>

$$r = -\frac{53}{12} - u s = -\frac{53}{12} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{35}{3}

$$(-\frac{53}{12} - u) (-\frac{53}{12} + u) = -\frac{35}{3}$$

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

$$\frac{2809}{144} - u^2 = -\frac{35}{3}$$

Simplify the expression by subtracting \frac{2809}{144} on both sides

$$-u^2 = -\frac{35}{3}-\frac{2809}{144} = -\frac{4489}{144}$$

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

$$u^2 = \frac{4489}{144} u = \pm\sqrt{\frac{4489}{144}} = \pm \frac{67}{12}$$

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

$$r =-\frac{53}{12} - \frac{67}{12} = -10 s = -\frac{53}{12} + \frac{67}{12} = 1.167$$

Evaluate

$$\left(6r-7\right)\left(r+10\right)$$