Solve 21c^2+11c-2

$$21 c ^ { 2 } + 11 c - 2$$

Factor

$$\left(7c-1\right)\left(3c+2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 21c^{2}+ac+bc-2. To find a and b, set up a system to be solved.

$$a+b=11 ab=21\left(-2\right)=-42$$

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 11.

$$a=-3 b=14$$

Rewrite 21c^{2}+11c-2 as \left(21c^{2}-3c\right)+\left(14c-2\right).

$$\left(21c^{2}-3c\right)+\left(14c-2\right)$$

Factor out 3c in the first and 2 in the second group.

$$3c\left(7c-1\right)+2\left(7c-1\right)$$

Factor out common term 7c-1 by using distributive property.

$$\left(7c-1\right)\left(3c+2\right)$$

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.

$$21c^{2}+11c-2=0$$

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.

$$c=\frac{-11±\sqrt{11^{2}-4\times 21\left(-2\right)}}{2\times 21}$$

Square 11.

$$c=\frac{-11±\sqrt{121-4\times 21\left(-2\right)}}{2\times 21}$$

Multiply -4 times 21.

$$c=\frac{-11±\sqrt{121-84\left(-2\right)}}{2\times 21}$$

Multiply -84 times -2.

$$c=\frac{-11±\sqrt{121+168}}{2\times 21}$$

Add 121 to 168.

$$c=\frac{-11±\sqrt{289}}{2\times 21}$$

Take the square root of 289.

$$c=\frac{-11±17}{2\times 21}$$

Multiply 2 times 21.

$$c=\frac{-11±17}{42}$$

Now solve the equation c=\frac{-11±17}{42} when ± is plus. Add -11 to 17.

$$c=\frac{6}{42}$$

Reduce the fraction \frac{6}{42} to lowest terms by extracting and canceling out 6.

$$c=\frac{1}{7}$$

Now solve the equation c=\frac{-11±17}{42} when ± is minus. Subtract 17 from -11.

$$c=-\frac{28}{42}$$

Reduce the fraction \frac{-28}{42} to lowest terms by extracting and canceling out 14.

$$c=-\frac{2}{3}$$

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

$$21c^{2}+11c-2=21\left(c-\frac{1}{7}\right)\left(c-\left(-\frac{2}{3}\right)\right)$$

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

$$21c^{2}+11c-2=21\left(c-\frac{1}{7}\right)\left(c+\frac{2}{3}\right)$$

Subtract \frac{1}{7} from c by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$21c^{2}+11c-2=21\times \frac{7c-1}{7}\left(c+\frac{2}{3}\right)$$

Add \frac{2}{3} to c by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$21c^{2}+11c-2=21\times \frac{7c-1}{7}\times \frac{3c+2}{3}$$

Multiply \frac{7c-1}{7} times \frac{3c+2}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$21c^{2}+11c-2=21\times \frac{\left(7c-1\right)\left(3c+2\right)}{7\times 3}$$

Multiply 7 times 3.

$$21c^{2}+11c-2=21\times \frac{\left(7c-1\right)\left(3c+2\right)}{21}$$

Cancel out 21, the greatest common factor in 21 and 21.

$$21c^{2}+11c-2=\left(7c-1\right)\left(3c+2\right)$$

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 21

$$x ^ 2 +\frac{11}{21}x -\frac{2}{21} = 0$$

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

$$r + s = -\frac{11}{21} rs = -\frac{2}{21}$$

Two numbers r and s sum up to -\frac{11}{21} exactly when the average of the two numbers is \frac{1}{2}*-\frac{11}{21} = -\frac{11}{42}. 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>

$$r = -\frac{11}{42} - u s = -\frac{11}{42} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{2}{21}

$$(-\frac{11}{42} - u) (-\frac{11}{42} + u) = -\frac{2}{21}$$

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

$$\frac{121}{1764} - u^2 = -\frac{2}{21}$$

Simplify the expression by subtracting \frac{121}{1764} on both sides

$$-u^2 = -\frac{2}{21}-\frac{121}{1764} = -\frac{289}{1764}$$

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

$$u^2 = \frac{289}{1764} u = \pm\sqrt{\frac{289}{1764}} = \pm \frac{17}{42}$$

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

$$r =-\frac{11}{42} - \frac{17}{42} = -0.667 s = -\frac{11}{42} + \frac{17}{42} = 0.143$$

Evaluate

$$\left(7c-1\right)\left(3c+2\right)$$