Solve 81x^2+90x+25

$$81 { x }^{ 2 } +90x+25$$

Factor

$$\left(9x+5\right)^{2}$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 81x^{2}+ax+bx+25. To find a and b, set up a system to be solved.

$$a+b=90 ab=81\times 25=2025$$

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 2025.

$$1,2025 3,675 5,405 9,225 15,135 25,81 27,75 45,45$$

Calculate the sum for each pair.

$$1+2025=2026 3+675=678 5+405=410 9+225=234 15+135=150 25+81=106 27+75=102 45+45=90$$

The solution is the pair that gives sum 90.

$$a=45 b=45$$

Rewrite 81x^{2}+90x+25 as \left(81x^{2}+45x\right)+\left(45x+25\right).

$$\left(81x^{2}+45x\right)+\left(45x+25\right)$$

Factor out 9x in the first and 5 in the second group.

$$9x\left(9x+5\right)+5\left(9x+5\right)$$

Factor out common term 9x+5 by using distributive property.

$$\left(9x+5\right)\left(9x+5\right)$$

Rewrite as a binomial square.

$$\left(9x+5\right)^{2}$$

Steps Using Square Of Binomial

This trinomial has the form of a trinomial square, perhaps multiplied by a common factor. Trinomial squares can be factored by finding the square roots of the leading and trailing terms.

$$factor(81x^{2}+90x+25)$$

Find the greatest common factor of the coefficients.

$$gcf(81,90,25)=1$$

Find the square root of the leading term, 81x^{2}.

$$\sqrt{81x^{2}}=9x$$

Find the square root of the trailing term, 25.

$$\sqrt{25}=5$$

The trinomial square is the square of the binomial that is the sum or difference of the square roots of the leading and trailing terms, with the sign determined by the sign of the middle term of the trinomial square.

$$\left(9x+5\right)^{2}$$

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.

$$81x^{2}+90x+25=0$$

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.

$$x=\frac{-90±\sqrt{90^{2}-4\times 81\times 25}}{2\times 81}$$

Square 90.

$$x=\frac{-90±\sqrt{8100-4\times 81\times 25}}{2\times 81}$$

Multiply -4 times 81.

$$x=\frac{-90±\sqrt{8100-324\times 25}}{2\times 81}$$

Multiply -324 times 25.

$$x=\frac{-90±\sqrt{8100-8100}}{2\times 81}$$

Add 8100 to -8100.

$$x=\frac{-90±\sqrt{0}}{2\times 81}$$

Take the square root of 0.

$$x=\frac{-90±0}{2\times 81}$$

Multiply 2 times 81.

$$x=\frac{-90±0}{162}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{5}{9} for x_{1} and -\frac{5}{9} for x_{2}.

$$81x^{2}+90x+25=81\left(x-\left(-\frac{5}{9}\right)\right)\left(x-\left(-\frac{5}{9}\right)\right)$$

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

$$81x^{2}+90x+25=81\left(x+\frac{5}{9}\right)\left(x+\frac{5}{9}\right)$$

Add \frac{5}{9} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$81x^{2}+90x+25=81\times \frac{9x+5}{9}\left(x+\frac{5}{9}\right)$$

Add \frac{5}{9} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$81x^{2}+90x+25=81\times \frac{9x+5}{9}\times \frac{9x+5}{9}$$

Multiply \frac{9x+5}{9} times \frac{9x+5}{9} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$81x^{2}+90x+25=81\times \frac{\left(9x+5\right)\left(9x+5\right)}{9\times 9}$$

Multiply 9 times 9.

$$81x^{2}+90x+25=81\times \frac{\left(9x+5\right)\left(9x+5\right)}{81}$$

Cancel out 81, the greatest common factor in 81 and 81.

$$81x^{2}+90x+25=\left(9x+5\right)\left(9x+5\right)$$

Evaluate

$$\left(9x+5\right)^{2}$$