Solve 21x^2+55x+36

$$21 { x }^{ 2 } +55x+36$$

Factor

$$\left(3x+4\right)\left(7x+9\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 21x^{2}+ax+bx+36. To find a and b, set up a system to be solved.

$$a+b=55 ab=21\times 36=756$$

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 756.

$$1,756 2,378 3,252 4,189 6,126 7,108 9,84 12,63 14,54 18,42 21,36 27,28$$

Calculate the sum for each pair.

$$1+756=757 2+378=380 3+252=255 4+189=193 6+126=132 7+108=115 9+84=93 12+63=75 14+54=68 18+42=60 21+36=57 27+28=55$$

The solution is the pair that gives sum 55.

$$a=27 b=28$$

Rewrite 21x^{2}+55x+36 as \left(21x^{2}+27x\right)+\left(28x+36\right).

$$\left(21x^{2}+27x\right)+\left(28x+36\right)$$

Factor out 3x in the first and 4 in the second group.

$$3x\left(7x+9\right)+4\left(7x+9\right)$$

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

$$\left(7x+9\right)\left(3x+4\right)$$

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.

$$21x^{2}+55x+36=0$$

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.

$$x=\frac{-55±\sqrt{55^{2}-4\times 21\times 36}}{2\times 21}$$

Square 55.

$$x=\frac{-55±\sqrt{3025-4\times 21\times 36}}{2\times 21}$$

Multiply -4 times 21.

$$x=\frac{-55±\sqrt{3025-84\times 36}}{2\times 21}$$

Multiply -84 times 36.

$$x=\frac{-55±\sqrt{3025-3024}}{2\times 21}$$

Add 3025 to -3024.

$$x=\frac{-55±\sqrt{1}}{2\times 21}$$

Take the square root of 1.

$$x=\frac{-55±1}{2\times 21}$$

Multiply 2 times 21.

$$x=\frac{-55±1}{42}$$

Now solve the equation x=\frac{-55±1}{42} when ± is plus. Add -55 to 1.

$$x=-\frac{54}{42}$$

Reduce the fraction \frac{-54}{42} to lowest terms by extracting and canceling out 6.

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

Now solve the equation x=\frac{-55±1}{42} when ± is minus. Subtract 1 from -55.

$$x=-\frac{56}{42}$$

Reduce the fraction \frac{-56}{42} to lowest terms by extracting and canceling out 14.

$$x=-\frac{4}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{9}{7} for x_{1} and -\frac{4}{3} for x_{2}.

$$21x^{2}+55x+36=21\left(x-\left(-\frac{9}{7}\right)\right)\left(x-\left(-\frac{4}{3}\right)\right)$$

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

$$21x^{2}+55x+36=21\left(x+\frac{9}{7}\right)\left(x+\frac{4}{3}\right)$$

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

$$21x^{2}+55x+36=21\times \frac{7x+9}{7}\left(x+\frac{4}{3}\right)$$

Add \frac{4}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$21x^{2}+55x+36=21\times \frac{7x+9}{7}\times \frac{3x+4}{3}$$

Multiply \frac{7x+9}{7} times \frac{3x+4}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$21x^{2}+55x+36=21\times \frac{\left(7x+9\right)\left(3x+4\right)}{7\times 3}$$

Multiply 7 times 3.

$$21x^{2}+55x+36=21\times \frac{\left(7x+9\right)\left(3x+4\right)}{21}$$

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

$$21x^{2}+55x+36=\left(7x+9\right)\left(3x+4\right)$$

Evaluate

$$\left(3x+4\right)\left(7x+9\right)$$