Solve 18x^2+41x+21

$$18 x ^ { 2 } + 41 x + 21$$

Factor

$$\left(2x+3\right)\left(9x+7\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 18x^{2}+ax+bx+21. To find a and b, set up a system to be solved.

$$a+b=41 ab=18\times 21=378$$

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 378.

$$1,378 2,189 3,126 6,63 7,54 9,42 14,27 18,21$$

Calculate the sum for each pair.

$$1+378=379 2+189=191 3+126=129 6+63=69 7+54=61 9+42=51 14+27=41 18+21=39$$

The solution is the pair that gives sum 41.

$$a=14 b=27$$

Rewrite 18x^{2}+41x+21 as \left(18x^{2}+14x\right)+\left(27x+21\right).

$$\left(18x^{2}+14x\right)+\left(27x+21\right)$$

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

$$2x\left(9x+7\right)+3\left(9x+7\right)$$

Factor out common term 9x+7 by using distributive property.

$$\left(9x+7\right)\left(2x+3\right)$$

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.

$$18x^{2}+41x+21=0$$

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.

$$x=\frac{-41±\sqrt{41^{2}-4\times 18\times 21}}{2\times 18}$$

Square 41.

$$x=\frac{-41±\sqrt{1681-4\times 18\times 21}}{2\times 18}$$

Multiply -4 times 18.

$$x=\frac{-41±\sqrt{1681-72\times 21}}{2\times 18}$$

Multiply -72 times 21.

$$x=\frac{-41±\sqrt{1681-1512}}{2\times 18}$$

Add 1681 to -1512.

$$x=\frac{-41±\sqrt{169}}{2\times 18}$$

Take the square root of 169.

$$x=\frac{-41±13}{2\times 18}$$

Multiply 2 times 18.

$$x=\frac{-41±13}{36}$$

Now solve the equation x=\frac{-41±13}{36} when ± is plus. Add -41 to 13.

$$x=-\frac{28}{36}$$

Reduce the fraction \frac{-28}{36} to lowest terms by extracting and canceling out 4.

$$x=-\frac{7}{9}$$

Now solve the equation x=\frac{-41±13}{36} when ± is minus. Subtract 13 from -41.

$$x=-\frac{54}{36}$$

Reduce the fraction \frac{-54}{36} to lowest terms by extracting and canceling out 18.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{7}{9} for x_{1} and -\frac{3}{2} for x_{2}.

$$18x^{2}+41x+21=18\left(x-\left(-\frac{7}{9}\right)\right)\left(x-\left(-\frac{3}{2}\right)\right)$$

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

$$18x^{2}+41x+21=18\left(x+\frac{7}{9}\right)\left(x+\frac{3}{2}\right)$$

Add \frac{7}{9} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$18x^{2}+41x+21=18\times \frac{9x+7}{9}\left(x+\frac{3}{2}\right)$$

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

$$18x^{2}+41x+21=18\times \frac{9x+7}{9}\times \frac{2x+3}{2}$$

Multiply \frac{9x+7}{9} times \frac{2x+3}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$18x^{2}+41x+21=18\times \frac{\left(9x+7\right)\left(2x+3\right)}{9\times 2}$$

Multiply 9 times 2.

$$18x^{2}+41x+21=18\times \frac{\left(9x+7\right)\left(2x+3\right)}{18}$$

Cancel out 18, the greatest common factor in 18 and 18.

$$18x^{2}+41x+21=\left(9x+7\right)\left(2x+3\right)$$

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 18

$$x ^ 2 +\frac{41}{18}x +\frac{7}{6} = 0$$

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

$$r + s = -\frac{41}{18} rs = \frac{7}{6}$$

Two numbers r and s sum up to -\frac{41}{18} exactly when the average of the two numbers is \frac{1}{2}*-\frac{41}{18} = -\frac{41}{36}. 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>

$$r = -\frac{41}{36} - u s = -\frac{41}{36} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{7}{6}

$$(-\frac{41}{36} - u) (-\frac{41}{36} + u) = \frac{7}{6}$$

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

$$\frac{1681}{1296} - u^2 = \frac{7}{6}$$

Simplify the expression by subtracting \frac{1681}{1296} on both sides

$$-u^2 = \frac{7}{6}-\frac{1681}{1296} = -\frac{169}{1296}$$

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

$$u^2 = \frac{169}{1296} u = \pm\sqrt{\frac{169}{1296}} = \pm \frac{13}{36}$$

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

$$r =-\frac{41}{36} - \frac{13}{36} = -1.500 s = -\frac{41}{36} + \frac{13}{36} = -0.778$$

Evaluate

$$\left(2x+3\right)\left(9x+7\right)$$