Web Lesson 07: Binomial Expansions
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Binomial Expansions
To expand: (A + B)n
- Write out the numbers of the nth line of Pascal's triangle
- Then write decreasing powers of ‘A’ (starting with An), next to each of the numbers
- Next, write increasing powers of ‘B’ (starting with B0), next to each of the numbers
- Simplify carefully
- Check you answer is correct [by putting any number (say, x=1.2) into the original expression and into your expansion. The answers should be the same...]
A few extra examples will demonstrate the process for you...
e.g. 1: Expand (2 + 3x)3
- Write out the numbers of the 3rd line of Pascal's triangle:
1 + 3 + 3 + 1
- Insert powers of ' 2 ' [starting from 2³ and decreasing the power]:
1(2)3 + 3(2)2 + 3(2)1 + 1(2)0
- Insert powers of ' 3x ' [starting from (3x)0 and increasing the power]:
1(2)3(3x)0 + 3(2)2(3x)1 + 3(2)1(3x)2 + 1(2)0(3x)3
- Simplify [Remember (anything)0 = 1 ], [Remember (3x)2 = (3)2(x)2 = 9x2 ]
1(8)(1) + 3(4)(3x) + 3(2)(9x2) + 1(1)(27x3)
8 + 36x + 54x2 + 27x3
- Use your calculator to CHECK [by putting x = (say) 1.2 into ‘the binomial’ and ‘your expansion’]
To put x = 1.2 into: (2 + 3x)³: On the calculator, TYPE: 1.2 STO x
Then TYPE: ( 2 + 3 x ) x□ 3 =
That should give: "175.616"
Now, let's use the calculator to put x = 1.2: into: 8 + 36x + 54x2 + 27x3:
Then TYPE: 8 + 36 x + 54 x x² + 27 x x³ =
That should give: "175.616" ✔ correct
e.g. 2: Expand (3 - ½x)4
- Write out the numbers of the 4th line of Pascal's triangle:
1 + 4 + 6 + 4 + 1
- Insert powers of ' 3 ' [starting from 34 and decreasing the power]:
1(3)4 + 4(3)3 + 6(3)2 + 4(3)1 + 1(3)0
- Insert powers of ' -½x ' [starting from (-½x)0 and increasing the power]:
1(3)4(-½x)0 + 4(3)3(-½x)1 + 6(3)2(-½x)2 + 4(3)1(-½x)3 + 1(3)0(-½x)4
- Simplify [Remember (anything)0 = 1 ], [Remember (-½x)2 = (-½)2(x)2 = (+¼)(x²) = ¼x2 ]
1(81)(1) + 4(27)(-½x) + 6(9)(¼x2) + 4(3)(-⅛x3) + 1(1)(1x4/16)
81 - 54x + 27x2 - 3x3 + 1x4
2 2 16
- Check
To put x = 1.2 into: (3 - ½x)4: On the calculator, TYPE: 1.2 STO x
Then TYPE: ( 3 - ½ x ) x□ 4 =
That gives: "………"
Now, use the calculator to put x = 1.2: into: 81 - 54x + ²⁷⁄₂ x2 - ³⁄₂ x3 + ¹⁄₁₆ x4:
Then TYPE: 81 - 54 x + □⁄□ ²⁷⁄₂ x x² - □⁄□ ³⁄₂ x x³ + □⁄□ ¹⁄₁₆ x x□ 4 =
That should give: "33.1776" ✔ correct
Question 1: Expand (1 + 2x)5
Clue:
1. Write out the numbers of the 5th line of Pascal's triangle:
1 + 5 + 10 + 10 + 5 + 1
2. Insert powers of ' 1 ' [starting from 15 and decreasing the power]:
1(1)5 + 5(1)4 + 10(1)3 + 10(1)2 + 5(1)1 + 1(1)0
Note: We might as well IGNORE these ‘Powers of 1’ as any power of 1 gives: ‘1’
1 + 5 + 10 + 10 + 5 + 1
3. Insert powers of '2x' [starting from (2x)0 and increasing the power]:
1(2x)0 + 5(2x)1 + 10(2x)… + 10(2x)… + 5(2x)… + 1(2x)5
4. Simplify [Remember (2x)3 = (2)3(x)3 = 8x3]
1(1) + 5(…) + 10(4x2) + 10(…x3) + 5(…x4) + 1(32x5)
1 + …x + …x2 + 80x3 + …x4 + …x5
5. Check:
To put x = 1.2 into: (1 + 2x)5: On the calculator, TYPE: 1.2 STO x
Then TYPE: ( 1 + 2 x ) x□ 5 =
That should give: "454.35424"
Now, let's use the calculator to put x = 1.2: into: 8 + 36x + 54x2 + 27x3:
Then TYPE:
1 + …… x + …… x x² + 80 x x³ …… x x□ 4 …… x x□ 5 =
That should give: "454.35424" ✔ correct
Question 2: Expand (1 + 2x)4
Clue:
Use the same method as you did in question 1 above, except this time, using the 4th line of
Pascal's triangle…
Again, this is easier than the examples in the movie and the 2 examples at the start of this
Web Lesson - because the ‘Powers of 1’ all just give ‘1’…
The correct answer is: 1 + 8x + …x2 + 32x3 + 16x4
Question 3: Expand (2 - ½x)3
Clue:
1. Write out the numbers of the 3rd line of Pascal's triangle
1 + 3 + 3 + 1
2. Insert powers of '2' [starting from 23 and decreasing the power]
1(2)3 + 3(2)… + 3(2)… + 1(2)…
3. Insert powers of '-½x' [starting from (-½x)0 and increasing the power]:
1(2)3(-½x)0 + 3(…)2(…)1 + 3(2)…(-½x)… + 1(2)0(…)…
4. Simplify
1(8)(1) + 3(…)(-½x) + 3(2)(¼x2) + 1(1)(-⅛x3)
8 - …x + …x² - …x³
5. Check
On the calculator, TYPE: 1.2 STO x
Then type: (2 - ½x)3 on the calculator and note the answer
Lastly, type: 8 - …x + …x2 - …x3 and see if it gives the exact same answer?
Question 4: Expand: (1 + 2x)3
Clue:
1. Write out the numbers of the 3rd line of Pascal's triangle
1 + 3 + 3 + 1
2. Insert powers of '1' [starting from 13 and decreasing the power]
1(1)3 + 3(1)… + 3(1)… + 1(1)…
=> 1 + 3 + 3 + 1
3. Insert powers of '2x' [starting from (2x)0 and increasing the power]:
1(2x)0 + 3(…)1 + 3(2x)… + 1(…)…
4. Simplify
1(1) + 3(2x) + 3(4x2) + 1(…)
1 + 6x + …x² + …x³
5. Check
On the calculator, TYPE: 1.2 STO x
Then type: (1 + 2x)3 = and note the answer…
Lastly, type: 1 + 6x + …x2 + …x3 and see if it gives the same answer?
Question 5: Expand: (1 + ½x)4
Clue:
Use the same method as you did in question 4 above, except this time, using the 4th line of
Pascal's triangle and using ‘½x’ instead of ‘2x’…
The correct answer is: 1 + 2x + …x2 + ½x3 + …x4
Question 6: Expand: (½ + ¾x)4
Clue:
1. Write out the numbers of the 4th line of Pascal's triangle
1 + 4 + 6 + 4 + 1
2. Insert powers of '½' [starting from ½4 and decreasing the power]
1(½)4 + 4(½)… + 6(½)… + 4(½)… + 1(½)0
3. Insert powers of '¾x' [starting from (¾x)0 and increasing the power]:
1(½)4(¾x)0 + 4(…)3(…)1 + 6(½)…(¾x)… + 4(½)1(…)… + 1(½)0(¾x)4
4. Simplify
(½)4 = 1/16 (¾x)0 = 1x
(½)3 = … (¾x)1 = …
(½)2 = (¾x)2 = 9x²
16
(½)1 = 1/2 (¾x)3 = 27x³
81
(½)0 = 1 (¾x)4 = …
5. Check
On the calculator, TYPE: 1.2 STO x
Putting x = 1.2 into: (½ + ¾x)3 gives: …
Putting x = 1.2 into: 1/16 + …x + …x2 + …x3 gives: …
Question 7: Expand: (2x - 1)6
Clue: Sorry, no help for this one…
Question 8: Expand: (1 + x²)3
Clue:
1. Write out the numbers of the 3rd line of Pascal's triangle
2. Insert powers of ‘1’ [starting from 13 and decreasing the power]
Note: You could just ignore these as all ‘powers of 1’ are equal to ‘1’
3. Insert powers of x² [starting from (x²)0 and increasing the power]:
1(1)3(x²)0 + 3(1)2(x²)1 + 3(1)1(x²)2 + 1(1)0(x²)3
4. Simplify
1(1) + 3(x2) + 3(x4) + 1(…)
1 + 3x2 + ……… + ………
5. Check:
On the calculator, TYPE: 1.2 STO x
Putting x = 1.2 into: (1 + x²)5 gives: …
Putting x = 1.2 into: 1 + 3x2 + … + …… gives: …
Notice how, unlike the expansions we've done so far, the powers of ‘x’ in this expansion
rise in steps of ‘2’?
Question 9: Expand: (x + 1/x)4
Clue: 1(x)4(1/x)0 + 4(x)3(1/x)1 + 6(x)2(1/x)2 + 4(x)1(…)3 + 1(…)0(…)4
Question 10: Expand: (2 + x)(2 - x)³
Clue:
This question is a lot harder!
First expand (2-x)3 using Pascal's method to get: 8 - 12x + …… - ……
Then we need to multiply out: (2 + x)(8 - 12x + …… - ……)
Which involves a lot of arrows (8 to be precise) and some care in the simplifying
┌──┬────┬─────┬─────┐
│ ▼ ▼ ▼ ▼
(2 + x)(8 - 12x + … - …) = 16 - 24x + 12x² - …
│ ▲ ▲ ▲ ▲ 8x - 12x² + … - … +
└──────┴────┴─────┴─────┘
-------------------------------
16 - 16x + 0x² + … - …
Check: When I put in x = 1.2 into (2 + x)(2 - x)³, I get: ‘1.6384’
Do you get the same when you put x = 1.2 into your answer?
The pass mark (to avoid additional homework on this topic) is 8/10
Hand in your workings and answers
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