Question

1. Use Pascal's Triangle to expand the binomial.

(d – 3)6
d6 – 18d5 + 135d4 – 540d3 + 1,215d2 – 1,458d + 729
d6 + 18d5 + 135d4 + 540d3 + 1,215d2 + 1,458d + 729
d6 – 6d5 + 15d4 – 20d3 + 15d2 – 6d + 1
d6 + 6d5 + 15d4 + 20d3 + 15d2 + 6d + 1
2. Use the Binomial Theorem to expand the binomial. (3v + s)5
s5 – 5s4v + 10s3v2 – 10s2v3 + 5sv4 – v5
s5 + 15s4v + 90s3v2 + 270s2v3 + 405sv4 + 243v5
s5 + 45s4v + 270s3v2 + 810s2v3 + 1,215sv4 + 729v5
s5 + 15s4 + 90s3 + 270s2 + 405s + 243
3. What is the fourth term of (d – 4b)3?
b3
–b3
64b3
–64b3

Answer 1

1) option:A

2)
`(3v+s)^5=243v^5+405v^4s+270v^3s^2+90v^2s^3+15vs^4+s^5`

3)
`-64b^3`

Explanation:

The binomial expansion of
`(ax+by)^n` is given by:

`(ax+by)^n=\binom{n}{0} ax^(n-1) (by)^0+\binom{n}{1} (ax)^(n-1)(by)^(1)+\binom{n}{1}(ax)^(n-2)(by)^(2)+.......+\binom{n}{n}(ax)^(0)(by)^(n)`

1) on finding the expansion of
`(d-3)^6`.

`((d-3)^6)=(d^3-27-9d^2+27d)^2=d^6-18d^5+135d^4-540d^3+1215d^2-1458d+729`

Hence option A is correct.

2) on finding the expansion of

`(3v+s)^5`

Hence,

`(3v+s)^5=243v^5+405v^4s+270v^3s^2+90v^2s^3+15vs^4+s^5`

3) We need to find the fourth term of
`(d-4b)^3` i.e.the term of
`b^3`.

`(d-4b)^3=d^3+(-4b)^3+3d^2*(-4b)+3d*\\(-4b)^2\\(d-4b)^3=d^3-64b^3-12d^2b+48db^2`

hence, the fourth term is:
`-64b^3`

Answer 2

The correct answers are:

1) d⁶-18d⁵+135d⁴-540d³+1215d²-1458d+729

2) 243v⁵+405v⁴s+270v³s²+90v²s³+15vs⁴+s⁵

3) -64b³

Explanation:

We first write out Pascal's Triangle. We start with 1; below this is a row of two 1's; from this line down, we start with 1 and each successive term is the sum of the two above it. I have attached a picture of Pascal's Triangle.

The coefficients of our expanded polynomial are the numbers in each row of the triangle. The first row represents the coefficients of a binomial to the 0 power; the second row represents the 1st power; the third row, the 2nd power; etc.

Using this, the coefficients of a binomial to the 6th power are: 1, 6, 15, 20, 15, 6, 1.

As we expand our polynomial, our first term will be the first coefficient multiplied by the first term of the binomial to the 6th power and the second term to the 0 power. The second term will be the second coefficient multiplied by the first term of the binomial to the 5th power and the second term to the 1st power; the third term will be the third coefficient multiplied by the first term of the binomial to the 4th power and the second term to the 2nd power; etc. In each term of the expansion, the first term of the binomial loses a power while the second term gains a power. This gives us:

1d⁶(-3)⁰+6(d⁵(-3)¹)+15(d⁴(-3)²)+20(d³(-3)³)+15(d²(-3)⁴)+6(d¹(-3)⁵)+1(d⁰(-3)⁶)

= d⁶-18d⁵+15(9d⁴)+20(-27d³)+15(81d²)+6(-243d)+729

= d⁶-18d⁵+135d⁴-540d³+1215d²-1458d+729

2) The coefficients for the 5th power, based on Pascal's Triangle, would be:

1, 5, 10, 10, 5, 1

This gives us:

1(3v)⁵s⁰+5(3v)⁴s¹+10(3v)³s²+10(3v)²s³+5(3v)¹s⁴+1(3v)⁰s⁵

= 243v⁵+5(81v⁴s)+10(27v³s²)+10(9v²s³)+5(3vs⁴)+s⁵

= 243v⁵+405v⁴s+270v³s²+90v²s³+15vs⁴+s⁵

3) The coefficients of the third power in Pascal's Triangle would be:

1, 3, 3, 1

Since there are 4 terms in this row, the first term will have d³ and (-4b)⁰; in each successive term, d will lose an exponent and (-4b) will gain an exponent. In the fourth term, d will have an exponent of 0 and (-4b) will have an exponent of 3; this makes the fourth term

(-4b)³ = -64b³