Solve 21x^2+25x+6

$$21 x ^ { 2 } + 25 x + 6$$

Factor

$$\left(3x+1\right)\left(7x+6\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 21x^{2}+ax+bx+6. To find a and b, set up a system to be solved.

$$a+b=25 ab=21\times 6=126$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 126.

$$1,126 2,63 3,42 6,21 7,18 9,14$$

Calculate the sum for each pair.

$$1+126=127 2+63=65 3+42=45 6+21=27 7+18=25 9+14=23$$

The solution is the pair that gives sum 25.

$$a=7 b=18$$

Rewrite 21x^{2}+25x+6 as \left(21x^{2}+7x\right)+\left(18x+6\right).

$$\left(21x^{2}+7x\right)+\left(18x+6\right)$$

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

$$7x\left(3x+1\right)+6\left(3x+1\right)$$

Factor out common term 3x+1 by using distributive property.

$$\left(3x+1\right)\left(7x+6\right)$$

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.

$$21x^{2}+25x+6=0$$

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.

$$x=\frac{-25±\sqrt{25^{2}-4\times 21\times 6}}{2\times 21}$$

Square 25.

$$x=\frac{-25±\sqrt{625-4\times 21\times 6}}{2\times 21}$$

Multiply -4 times 21.

$$x=\frac{-25±\sqrt{625-84\times 6}}{2\times 21}$$

Multiply -84 times 6.

$$x=\frac{-25±\sqrt{625-504}}{2\times 21}$$

Add 625 to -504.

$$x=\frac{-25±\sqrt{121}}{2\times 21}$$

Take the square root of 121.

$$x=\frac{-25±11}{2\times 21}$$

Multiply 2 times 21.

$$x=\frac{-25±11}{42}$$

Now solve the equation x=\frac{-25±11}{42} when ± is plus. Add -25 to 11.

$$x=-\frac{14}{42}$$

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

$$x=-\frac{1}{3}$$

Now solve the equation x=\frac{-25±11}{42} when ± is minus. Subtract 11 from -25.

$$x=-\frac{36}{42}$$

Reduce the fraction \frac{-36}{42} to lowest terms by extracting and canceling out 6.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{1}{3} for x_{1} and -\frac{6}{7} for x_{2}.

$$21x^{2}+25x+6=21\left(x-\left(-\frac{1}{3}\right)\right)\left(x-\left(-\frac{6}{7}\right)\right)$$

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

$$21x^{2}+25x+6=21\left(x+\frac{1}{3}\right)\left(x+\frac{6}{7}\right)$$

Add \frac{1}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$21x^{2}+25x+6=21\times \frac{3x+1}{3}\left(x+\frac{6}{7}\right)$$

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

$$21x^{2}+25x+6=21\times \frac{3x+1}{3}\times \frac{7x+6}{7}$$

Multiply \frac{3x+1}{3} times \frac{7x+6}{7} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$21x^{2}+25x+6=21\times \frac{\left(3x+1\right)\left(7x+6\right)}{3\times 7}$$

Multiply 3 times 7.

$$21x^{2}+25x+6=21\times \frac{\left(3x+1\right)\left(7x+6\right)}{21}$$

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

$$21x^{2}+25x+6=\left(3x+1\right)\left(7x+6\right)$$

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

$$x ^ 2 +\frac{25}{21}x +\frac{2}{7} = 0$$

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

$$r + s = -\frac{25}{21} rs = \frac{2}{7}$$

Two numbers r and s sum up to -\frac{25}{21} exactly when the average of the two numbers is \frac{1}{2}*-\frac{25}{21} = -\frac{25}{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>

$$r = -\frac{25}{42} - u s = -\frac{25}{42} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{2}{7}

$$(-\frac{25}{42} - u) (-\frac{25}{42} + u) = \frac{2}{7}$$

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

$$\frac{625}{1764} - u^2 = \frac{2}{7}$$

Simplify the expression by subtracting \frac{625}{1764} on both sides

$$-u^2 = \frac{2}{7}-\frac{625}{1764} = -\frac{121}{1764}$$

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

$$u^2 = \frac{121}{1764} u = \pm\sqrt{\frac{121}{1764}} = \pm \frac{11}{42}$$

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

$$r =-\frac{25}{42} - \frac{11}{42} = -0.857 s = -\frac{25}{42} + \frac{11}{42} = -0.333$$

Evaluate

$$\left(3x+1\right)\left(7x+6\right)$$