Video Transcript
Use Pascal’s triangle to find the binomial expansion of 𝑥 plus 𝑦 to the fourth power.
Pascal’s triangle gives us the coefficients for an expanded binomial in the form 𝑎 plus 𝑏 to the 𝑛th power, where 𝑛 corresponds to the row of the triangle and begins with zero, which means the first row of the triangle is equal to 𝑛 equal zero. The second row corresponds to 𝑛 equals one, and so on. If we look carefully at our expression, we are working with 𝑥 plus 𝑦 to the fourth power, which will be the fifth row in Pascal’s triangle.
And how it works is this way. Using the triangle, we take 𝑎 to the 𝑛th power multiplied by 𝑏 to the zero power all the way on the left. And as we move to the right, we subtract one from 𝑎’s exponents and add it to 𝑏’s so that we have for our second term 𝑎 to the 𝑛 minus one power and 𝑏 to the first power. And we continue this pattern until 𝑎 is taken to the zero power and 𝑏 is taken to the 𝑛th power. And Pascal’s triangle tells us what the coefficients will be for each of these terms. Because we’re operating in the fifth row, we can expect that there will be five terms and the coefficients will be one, four, six, four, one. Let’s see if we can follow the pattern.
For us, we’ll start with 𝑥 to the fourth power, 𝑦 to the zero power, and a coefficient of one. After that, we subtract one from the 𝑥-exponent and add one to the 𝑦-exponent. And following along on the triangle, it has a coefficient of four. Again, we’ll need to subtract one from the 𝑥-exponent and add one to the 𝑦-exponent. Using the triangle, we see that there will be a coefficient of six so that our next term is six 𝑥 squared 𝑦 squared. Continuing on, we subtract one again from the 𝑥-exponent to get 𝑥 to the first power. And we add one to the 𝑦-exponent to get 𝑦 cubed. And the next coefficient along Pascal’s triangle will be a four.
At this point, we do have four terms. And we should expect only to have one final term, which is true because at this point we’ll have 𝑥 to the zero power and 𝑦 to the fourth power. And again, we get a coefficient of one on this end. So let’s check our pattern. According to Pascal’s triangle, we have coefficients one, four, six, four, one. And we have that one, four, six, four, one. We started with our 𝑥 to the fourth power 𝑦 to the zero power. And we ended with our 𝑥 to the zero power 𝑦 to the fourth power. However, we want to do some simplification here. Since our first term has a coefficient of one, we can rewrite this as simply 𝑥 to the fourth power because 𝑦 to the zero power is one. And that means one times 𝑥 to the fourth power times one just equals 𝑥 to the fourth power.
Our second term is then four times 𝑥 cubed times 𝑦. We don’t have to indicate 𝑦 to the first power as 𝑦 to the first power is equal to 𝑦. Then we’ll have six 𝑥 squared 𝑦 squared, which doesn’t simplify any further, plus four 𝑥𝑦 cubed. We don’t have to have 𝑥 to the first power as 𝑥 to the first power is equal to 𝑥. And finally, one times 𝑥 to the zero power equals one, which is multiplied by 𝑦 to the fourth power. So we can simplify that term to be 𝑦 to the fourth power. Using Pascal’s triangle, we’ve shown that 𝑥 plus 𝑦 to the fourth power is equal to 𝑥 to the fourth power plus four times 𝑥 cubed times 𝑦 plus six times 𝑥 squared 𝑦 squared plus four times 𝑥 times 𝑦 cubed plus 𝑦 to the fourth power.
