# How do you expand (2y^2+x)^7?

Oct 4, 2017

${\left(2 {y}^{2} + x\right)}^{7} = 128 {y}^{14} + 448 x {y}^{12} + 672 {x}^{2} {y}^{10} + 560 {x}^{3} {y}^{8} + 280 {x}^{4} {y}^{6} + 84 {x}^{5} {y}^{4} + 14 {x}^{6} {y}^{2} + {x}^{7}$

#### Explanation:

When expanding an expression like this, knowing Pascal's Triangle is crucial:

(Source)

Since this expression is of degree 7 (i.e. has highest power 7), there will be 8 terms in the expansion (like in a quadratic equation where the highest power is 2, there are 3 terms).

As you can see, the $8 t h$ row has six numbers $\left\{1 , 7 , 21 , 35 , 35 , 21 , 7 , 1\right\}$ respectively.

We use these numbers as coefficients in the expansion of ${\left(2 {y}^{2} + x\right)}^{7}$.

As you can see from the answer, the powers on the $2 y$ are descending as the powers on the $x$ ascend. The exponents of $2 y$ and $x$ must always add up to 7, the expression's highest power.

Basically, to expand this expression you need to multiply the numbers of Pascal's Triangle with the exponents on the first term of the expression ($2 y$), descending from the highest degree 7 to 0. Simultaneously, you need to do this same thing with the second term ($x$), but ascending from 0 to 7 instead:

${\left(2 {y}^{2} + x\right)}^{7}$

$= 1 {\left(2 {y}^{2}\right)}^{7} {\left(x\right)}^{0} + 7 {\left(2 {y}^{2}\right)}^{6} {\left(x\right)}^{1} + 21 {\left(2 {y}^{2}\right)}^{5} {\left(x\right)}^{2} + 35 {\left(2 {y}^{2}\right)}^{4} {\left(x\right)}^{3} + 35 {\left(2 {y}^{2}\right)}^{3} {\left(x\right)}^{4} + 21 {\left(2 {y}^{2}\right)}^{2} {\left(x\right)}^{5} + 7 {\left(2 {y}^{2}\right)}^{1} {\left(x\right)}^{6} + 1 {\left(2 {y}^{2}\right)}^{0} {\left(x\right)}^{7}$

$= 128 {y}^{14} + 448 x {y}^{12} + 672 {x}^{2} {y}^{10} + 560 {x}^{3} {y}^{8} + 280 {x}^{4} {y}^{6} + 84 {x}^{5} {y}^{4} + 14 {x}^{6} {y}^{2} + {x}^{7}$

(This response has been closely based off HSBC244's reply to a similar question).