Solve 35r^2+r-12=

$$35 r ^ { 2 } + r - 12 =$$

Factor

$$\left(7r-4\right)\left(5r+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 35r^{2}+ar+br-12. To find a and b, set up a system to be solved.

$$a+b=1 ab=35\left(-12\right)=-420$$

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 1.

$$a=-20 b=21$$

Rewrite 35r^{2}+r-12 as \left(35r^{2}-20r\right)+\left(21r-12\right).

$$\left(35r^{2}-20r\right)+\left(21r-12\right)$$

Factor out 5r in the first and 3 in the second group.

$$5r\left(7r-4\right)+3\left(7r-4\right)$$

Factor out common term 7r-4 by using distributive property.

$$\left(7r-4\right)\left(5r+3\right)$$

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.

$$35r^{2}+r-12=0$$

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.

$$r=\frac{-1±\sqrt{1^{2}-4\times 35\left(-12\right)}}{2\times 35}$$

Square 1.

$$r=\frac{-1±\sqrt{1-4\times 35\left(-12\right)}}{2\times 35}$$

Multiply -4 times 35.

$$r=\frac{-1±\sqrt{1-140\left(-12\right)}}{2\times 35}$$

Multiply -140 times -12.

$$r=\frac{-1±\sqrt{1+1680}}{2\times 35}$$

Add 1 to 1680.

$$r=\frac{-1±\sqrt{1681}}{2\times 35}$$

Take the square root of 1681.

$$r=\frac{-1±41}{2\times 35}$$

Multiply 2 times 35.

$$r=\frac{-1±41}{70}$$

Now solve the equation r=\frac{-1±41}{70} when ± is plus. Add -1 to 41.

$$r=\frac{40}{70}$$

Reduce the fraction \frac{40}{70} to lowest terms by extracting and canceling out 10.

$$r=\frac{4}{7}$$

Now solve the equation r=\frac{-1±41}{70} when ± is minus. Subtract 41 from -1.

$$r=-\frac{42}{70}$$

Reduce the fraction \frac{-42}{70} to lowest terms by extracting and canceling out 14.

$$r=-\frac{3}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{4}{7} for x_{1} and -\frac{3}{5} for x_{2}.

$$35r^{2}+r-12=35\left(r-\frac{4}{7}\right)\left(r-\left(-\frac{3}{5}\right)\right)$$

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

$$35r^{2}+r-12=35\left(r-\frac{4}{7}\right)\left(r+\frac{3}{5}\right)$$

Subtract \frac{4}{7} from r by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$35r^{2}+r-12=35\times \frac{7r-4}{7}\left(r+\frac{3}{5}\right)$$

Add \frac{3}{5} to r by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$35r^{2}+r-12=35\times \frac{7r-4}{7}\times \frac{5r+3}{5}$$

Multiply \frac{7r-4}{7} times \frac{5r+3}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$35r^{2}+r-12=35\times \frac{\left(7r-4\right)\left(5r+3\right)}{7\times 5}$$

Multiply 7 times 5.

$$35r^{2}+r-12=35\times \frac{\left(7r-4\right)\left(5r+3\right)}{35}$$

Cancel out 35, the greatest common factor in 35 and 35.

$$35r^{2}+r-12=\left(7r-4\right)\left(5r+3\right)$$

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 35

$$x ^ 2 +\frac{1}{35}x -\frac{12}{35} = 0$$

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

$$r + s = -\frac{1}{35} rs = -\frac{12}{35}$$

Two numbers r and s sum up to -\frac{1}{35} exactly when the average of the two numbers is \frac{1}{2}*-\frac{1}{35} = -\frac{1}{70}. 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>

$$r = -\frac{1}{70} - u s = -\frac{1}{70} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{12}{35}

$$(-\frac{1}{70} - u) (-\frac{1}{70} + u) = -\frac{12}{35}$$

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

$$\frac{1}{4900} - u^2 = -\frac{12}{35}$$

Simplify the expression by subtracting \frac{1}{4900} on both sides

$$-u^2 = -\frac{12}{35}-\frac{1}{4900} = -\frac{1681}{4900}$$

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

$$u^2 = \frac{1681}{4900} u = \pm\sqrt{\frac{1681}{4900}} = \pm \frac{41}{70}$$

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

$$r =-\frac{1}{70} - \frac{41}{70} = -0.600 s = -\frac{1}{70} + \frac{41}{70} = 0.571$$

Evaluate

$$35r^{2}+r-12$$