# Question Video: Finding Binomial Coefficients Mathematics

In the expansion of (𝑥² − (1/2))⁸ according to the descending powers of 𝑥, find the second term from the end.

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### Video Transcript

In the expansion of 𝑥 squared minus a half all raised to the eighth power according to the descending powers of 𝑥, find the second term from the end.

This question contains a binomial expression of the form 𝑎 plus 𝑏 all raised to the 𝑛th power. We recall that the binomial theorem states that this is equal to 𝑛 choose zero multiplied by 𝑎 to the 𝑛th power plus 𝑛 choose one multiplied by 𝑎 to the 𝑛 minus oneth power multiplied by 𝑏 to the first power, and so on.

We have highlighted the general term, where 𝑛 choose 𝑟 is equal to 𝑛 factorial divided by 𝑛 minus 𝑟 factorial multiplied by 𝑟 factorial. As we move term by term from left to right, the exponents or powers of 𝑎 decrease and the exponents or powers of 𝑏 increase. The expression in this question is 𝑥 squared minus a half all raised to the eighth power. This means that 𝑎 is equal to 𝑥 squared, 𝑏 is equal to negative a half, and 𝑛 is equal to eight.

We could write out the whole expansion. However, we’re only interested in the second term from the end. From the general expansion, this would be equal to eight choose seven multiplied by 𝑎 to the first power multiplied by 𝑏 to the seventh power. Substituting in our values of 𝑎 and 𝑏, we have eight choose seven multiplied by 𝑥 squared to the first power multiplied by negative one-half to the seventh power.

Eight choose seven is equal to eight factorial divided by one factorial multiplied by seven factorial. We recall that eight factorial can be rewritten as eight multiplied by seven factorial. And one factorial is equal to one. We can then cancel seven factorial on the numerator and denominator such that eight choose seven is equal to eight. 𝑥 squared to the first power is just 𝑥 squared. And as negative two to the seventh power is negative 128, negative a half to the seventh power is negative one over 128. As eight divides into 128 16 times, our expression simplifies to negative 𝑥 squared over 16. This is the second term from the end of the expansion 𝑥 squared minus a half all raised to the eighth power.