Video: Use Pascal’s Triangle to Expand the Expression

Use Pascal’s triangle to expand the expression (π‘₯ + (1/π‘₯))⁴.

03:57

Video Transcript

Use Pascal’s triangle to expand the expression π‘₯ plus one over π‘₯ to the power of four.

Pascal’s triangle can be used to determine the pattern of coefficients for any binomial expansion. As the name suggests, it is in the shape of a triangle. We calculate the next row of the triangle by adding two numbers from the row above as shown. One plus two is equal to three, and six plus four is equal to 10.

As our exponent or power is four, there will be five terms in the expansion. This means that we use the row one, four, six, four, and one. These numbers can also be found using the 𝑛 𝐢 π‘Ÿ button on the calculator. Four choose zero is equal to one. Four choose one is equal to four. Four choose two is equal to six, and so on.

If we let the two terms inside our parentheses be π‘Ž and 𝑏, then the first coefficient will be multiplied by π‘Ž to the power of four multiplied by 𝑏 to the power of zero. The second coefficient is multiplied by π‘Ž to the power of three multiplied by 𝑏 to the power of one. The powers or exponents of π‘Ž decrease by one each time, whereas the exponents of 𝑏 increase by one each time.

The sum of the exponents in each term is equal to four. Four plus zero is equal to four. Three plus one equals four. And two plus two is equal to four. Substituting in our values means that the first term is equal to one multiplied by π‘₯ to the power of four multiplied by one over π‘₯ to the power of zero. The second term is equal to four multiplied by π‘₯ cubed multiplied by one over π‘₯ to the power of one. The third, fourth, and fifth terms are as shown.

Anything to the power of zero is equal to one. So one over π‘₯ to the power of zero and π‘₯ to the power of zero are equal to one. One over π‘₯ squared is equal to one over π‘₯ squared. We can square the numerator and denominator. In the same way, one over π‘₯ cubed is equal to one over π‘₯ cubed. And the final term, one over π‘₯ to the power of four, is equal to one over π‘₯ to the power of four.

The first term simplifies to π‘₯ to the power of four. The second term simplifies to four π‘₯ cubed multiplied by one over π‘₯. At this point, we can cancel an π‘₯. So the term becomes four π‘₯ squared. The third term is six π‘₯ squared multiplied by one over π‘₯ squared. The π‘₯ squareds cancel. So we’re left with six multiplied by one, which is equal to six.

The next term is equal to four π‘₯ multiplied by one over π‘₯ cubed. Once again, we can cancel an π‘₯, leaving us with four over π‘₯ squared. The final term is equal to one over π‘₯ to the power of four.

The simplified version of the expansion of π‘₯ plus one over π‘₯ to the power of four is π‘₯ to the power of four plus four π‘₯ squared plus six plus four over π‘₯ squared plus one over π‘₯ to the power of four.

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