Solve 3r^2+r-14

$$3 r ^ { 2 } + r - 14$$

Factor

$$\left(r-2\right)\left(3r+7\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3r^{2}+ar+br-14. To find a and b, set up a system to be solved.

$$a+b=1 ab=3\left(-14\right)=-42$$

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 -42.

$$-1,42 -2,21 -3,14 -6,7$$

Calculate the sum for each pair.

$$-1+42=41 -2+21=19 -3+14=11 -6+7=1$$

The solution is the pair that gives sum 1.

$$a=-6 b=7$$

Rewrite 3r^{2}+r-14 as \left(3r^{2}-6r\right)+\left(7r-14\right).

$$\left(3r^{2}-6r\right)+\left(7r-14\right)$$

Factor out 3r in the first and 7 in the second group.

$$3r\left(r-2\right)+7\left(r-2\right)$$

Factor out common term r-2 by using distributive property.

$$\left(r-2\right)\left(3r+7\right)$$

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.

$$3r^{2}+r-14=0$$

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.

$$r=\frac{-1±\sqrt{1^{2}-4\times 3\left(-14\right)}}{2\times 3}$$

Square 1.

$$r=\frac{-1±\sqrt{1-4\times 3\left(-14\right)}}{2\times 3}$$

Multiply -4 times 3.

$$r=\frac{-1±\sqrt{1-12\left(-14\right)}}{2\times 3}$$

Multiply -12 times -14.

$$r=\frac{-1±\sqrt{1+168}}{2\times 3}$$

Add 1 to 168.

$$r=\frac{-1±\sqrt{169}}{2\times 3}$$

Take the square root of 169.

$$r=\frac{-1±13}{2\times 3}$$

Multiply 2 times 3.

$$r=\frac{-1±13}{6}$$

Now solve the equation r=\frac{-1±13}{6} when ± is plus. Add -1 to 13.

$$r=\frac{12}{6}$$

Divide 12 by 6.

$$r=2$$

Now solve the equation r=\frac{-1±13}{6} when ± is minus. Subtract 13 from -1.

$$r=-\frac{14}{6}$$

Reduce the fraction \frac{-14}{6} to lowest terms by extracting and canceling out 2.

$$r=-\frac{7}{3}$$

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

$$3r^{2}+r-14=3\left(r-2\right)\left(r-\left(-\frac{7}{3}\right)\right)$$

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

$$3r^{2}+r-14=3\left(r-2\right)\left(r+\frac{7}{3}\right)$$

Add \frac{7}{3} to r by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$3r^{2}+r-14=3\left(r-2\right)\times \frac{3r+7}{3}$$

Cancel out 3, the greatest common factor in 3 and 3.

$$3r^{2}+r-14=\left(r-2\right)\left(3r+7\right)$$

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 3

$$x ^ 2 +\frac{1}{3}x -\frac{14}{3} = 0$$

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

$$r + s = -\frac{1}{3} rs = -\frac{14}{3}$$

Two numbers r and s sum up to -\frac{1}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{1}{3} = -\frac{1}{6}. 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>

$$r = -\frac{1}{6} - u s = -\frac{1}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{14}{3}

$$(-\frac{1}{6} - u) (-\frac{1}{6} + u) = -\frac{14}{3}$$

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

$$\frac{1}{36} - u^2 = -\frac{14}{3}$$

Simplify the expression by subtracting \frac{1}{36} on both sides

$$-u^2 = -\frac{14}{3}-\frac{1}{36} = -\frac{169}{36}$$

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

$$u^2 = \frac{169}{36} u = \pm\sqrt{\frac{169}{36}} = \pm \frac{13}{6}$$

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

$$r =-\frac{1}{6} - \frac{13}{6} = -2.333 s = -\frac{1}{6} + \frac{13}{6} = 2.000$$

Evaluate

$$\left(r-2\right)\left(3r+7\right)$$