Solve 81x^2+180x+100

$$81 x ^ { 2 } + 180 x + 100$$

Factor

$$\left(9x+10\right)^{2}$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 81x^{2}+ax+bx+100. To find a and b, set up a system to be solved.

$$a+b=180 ab=81\times 100=8100$$

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 8100.

$$1,8100 2,4050 3,2700 4,2025 5,1620 6,1350 9,900 10,810 12,675 15,540 18,450 20,405 25,324 27,300 30,270 36,225 45,180 50,162 54,150 60,135 75,108 81,100 90,90$$

Calculate the sum for each pair.

$$1+8100=8101 2+4050=4052 3+2700=2703 4+2025=2029 5+1620=1625 6+1350=1356 9+900=909 10+810=820 12+675=687 15+540=555 18+450=468 20+405=425 25+324=349 27+300=327 30+270=300 36+225=261 45+180=225 50+162=212 54+150=204 60+135=195 75+108=183 81+100=181 90+90=180$$

The solution is the pair that gives sum 180.

$$a=90 b=90$$

Rewrite 81x^{2}+180x+100 as \left(81x^{2}+90x\right)+\left(90x+100\right).

$$\left(81x^{2}+90x\right)+\left(90x+100\right)$$

Factor out 9x in the first and 10 in the second group.

$$9x\left(9x+10\right)+10\left(9x+10\right)$$

Factor out common term 9x+10 by using distributive property.

$$\left(9x+10\right)\left(9x+10\right)$$

Rewrite as a binomial square.

$$\left(9x+10\right)^{2}$$

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.

$$factor(81x^{2}+180x+100)$$

Find the greatest common factor of the coefficients.

$$gcf(81,180,100)=1$$

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

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

Find the square root of the trailing term, 100.

$$\sqrt{100}=10$$

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.

$$\left(9x+10\right)^{2}$$

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.

$$81x^{2}+180x+100=0$$

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.

$$x=\frac{-180±\sqrt{180^{2}-4\times 81\times 100}}{2\times 81}$$

Square 180.

$$x=\frac{-180±\sqrt{32400-4\times 81\times 100}}{2\times 81}$$

Multiply -4 times 81.

$$x=\frac{-180±\sqrt{32400-324\times 100}}{2\times 81}$$

Multiply -324 times 100.

$$x=\frac{-180±\sqrt{32400-32400}}{2\times 81}$$

Add 32400 to -32400.

$$x=\frac{-180±\sqrt{0}}{2\times 81}$$

Take the square root of 0.

$$x=\frac{-180±0}{2\times 81}$$

Multiply 2 times 81.

$$x=\frac{-180±0}{162}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{10}{9} for x_{1} and -\frac{10}{9} for x_{2}.

$$81x^{2}+180x+100=81\left(x-\left(-\frac{10}{9}\right)\right)\left(x-\left(-\frac{10}{9}\right)\right)$$

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

$$81x^{2}+180x+100=81\left(x+\frac{10}{9}\right)\left(x+\frac{10}{9}\right)$$

Add \frac{10}{9} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$81x^{2}+180x+100=81\times \frac{9x+10}{9}\left(x+\frac{10}{9}\right)$$

Add \frac{10}{9} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$81x^{2}+180x+100=81\times \frac{9x+10}{9}\times \frac{9x+10}{9}$$

Multiply \frac{9x+10}{9} times \frac{9x+10}{9} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$81x^{2}+180x+100=81\times \frac{\left(9x+10\right)\left(9x+10\right)}{9\times 9}$$

Multiply 9 times 9.

$$81x^{2}+180x+100=81\times \frac{\left(9x+10\right)\left(9x+10\right)}{81}$$

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

$$81x^{2}+180x+100=\left(9x+10\right)\left(9x+10\right)$$

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 81

$$x ^ 2 +\frac{20}{9}x +\frac{100}{81} = 0$$

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

$$r + s = -\frac{20}{9} rs = \frac{100}{81}$$

Two numbers r and s sum up to -\frac{20}{9} exactly when the average of the two numbers is \frac{1}{2}*-\frac{20}{9} = -\frac{10}{9}. 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>

$$r = -\frac{10}{9} - u s = -\frac{10}{9} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{100}{81}

$$(-\frac{10}{9} - u) (-\frac{10}{9} + u) = \frac{100}{81}$$

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

$$\frac{100}{81} - u^2 = \frac{100}{81}$$

Simplify the expression by subtracting \frac{100}{81} on both sides

$$-u^2 = \frac{100}{81}-\frac{100}{81} = 0$$

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

$$u^2 = 0 u = 0$$

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

$$r = s = -\frac{10}{9} = -1.1111111260122724$$

Evaluate

$$\left(9x+10\right)^{2}$$