Solve 81w^2+144w+64quad1

$$81 w ^ { 2 } + 144 w + 64 \quad 1$$

Factor

$$\left(9w+8\right)^{2}$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 81w^{2}+aw+bw+64. To find a and b, set up a system to be solved.

$$a+b=144 ab=81\times 64=5184$$

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 5184.

$$1,5184 2,2592 3,1728 4,1296 6,864 8,648 9,576 12,432 16,324 18,288 24,216 27,192 32,162 36,144 48,108 54,96 64,81 72,72$$

Calculate the sum for each pair.

$$1+5184=5185 2+2592=2594 3+1728=1731 4+1296=1300 6+864=870 8+648=656 9+576=585 12+432=444 16+324=340 18+288=306 24+216=240 27+192=219 32+162=194 36+144=180 48+108=156 54+96=150 64+81=145 72+72=144$$

The solution is the pair that gives sum 144.

$$a=72 b=72$$

Rewrite 81w^{2}+144w+64 as \left(81w^{2}+72w\right)+\left(72w+64\right).

$$\left(81w^{2}+72w\right)+\left(72w+64\right)$$

Factor out 9w in the first and 8 in the second group.

$$9w\left(9w+8\right)+8\left(9w+8\right)$$

Factor out common term 9w+8 by using distributive property.

$$\left(9w+8\right)\left(9w+8\right)$$

Rewrite as a binomial square.

$$\left(9w+8\right)^{2}$$

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.

$$factor(81w^{2}+144w+64)$$

Find the greatest common factor of the coefficients.

$$gcf(81,144,64)=1$$

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

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

Find the square root of the trailing term, 64.

$$\sqrt{64}=8$$

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.

$$\left(9w+8\right)^{2}$$

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.

$$81w^{2}+144w+64=0$$

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.

$$w=\frac{-144±\sqrt{144^{2}-4\times 81\times 64}}{2\times 81}$$

Square 144.

$$w=\frac{-144±\sqrt{20736-4\times 81\times 64}}{2\times 81}$$

Multiply -4 times 81.

$$w=\frac{-144±\sqrt{20736-324\times 64}}{2\times 81}$$

Multiply -324 times 64.

$$w=\frac{-144±\sqrt{20736-20736}}{2\times 81}$$

Add 20736 to -20736.

$$w=\frac{-144±\sqrt{0}}{2\times 81}$$

Take the square root of 0.

$$w=\frac{-144±0}{2\times 81}$$

Multiply 2 times 81.

$$w=\frac{-144±0}{162}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{8}{9} for x_{1} and -\frac{8}{9} for x_{2}.

$$81w^{2}+144w+64=81\left(w-\left(-\frac{8}{9}\right)\right)\left(w-\left(-\frac{8}{9}\right)\right)$$

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

$$81w^{2}+144w+64=81\left(w+\frac{8}{9}\right)\left(w+\frac{8}{9}\right)$$

Add \frac{8}{9} to w by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$81w^{2}+144w+64=81\times \frac{9w+8}{9}\left(w+\frac{8}{9}\right)$$

Add \frac{8}{9} to w by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$81w^{2}+144w+64=81\times \frac{9w+8}{9}\times \frac{9w+8}{9}$$

Multiply \frac{9w+8}{9} times \frac{9w+8}{9} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$81w^{2}+144w+64=81\times \frac{\left(9w+8\right)\left(9w+8\right)}{9\times 9}$$

Multiply 9 times 9.

$$81w^{2}+144w+64=81\times \frac{\left(9w+8\right)\left(9w+8\right)}{81}$$

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

$$81w^{2}+144w+64=\left(9w+8\right)\left(9w+8\right)$$

Evaluate

$$\left(9w+8\right)^{2}$$

Solution Steps

Multiply 64 and 1 to get 64.

$$81w^{2}+144w+64$$