High School

**Question 1:**

Expand \((x - 2)^6\).

A. \(x^6 + 12x^5 + 60x^4 + 160x^3 + 240x^2 + 192x + 64\)
B. \(x^6 - 12x^5 - 60x^4 - 160x^3 - 240x^2 - 192x - 64\)
C. \(x^6 - 12x^5 + 60x^4 - 160x^3 + 240x^2 - 192x + 64\)
D. \(x^6 - 12x^5 + 60x^4 - 160x^3 + 240x^2 - 160x + 64\)

**Question 2:**

Expand \((x + 2y)^4\).

A. \(x^4 - 8x^3y + 24x^2y^2 - 32xy^3 + 16y^4\)
B. \(x^4 + 8x^3y + 24x^2y^2 + 32xy^3 + 16y^4\)
C. \(x^4 + 8x^3y + 28x^2y^2 + 32xy^3 + 16y^4\)
D. \(x^4 + 8x^3y + 24x^2y^2 + 34xy^3 + 16y^4\)

**Question 3:**

How many terms are in the expansion of \((a + b)^{20}\)?

A. 19
B. 20
C. 21
D. 22

**Question 4:**

Expand \((x - y)^4\).

A. \(x^4 - 5x^3y + 6x^2y^2 - 5xy^3 + y^4\)
B. \(x^4 - 4x^3y + 8x^2y^2 - 4xy^3 + y^4\)
C. \(x^4 - 4x^3y + 6x^2y^2 - 4xy^3 + y^4\)
D. \(x^4 - 4x^3y + 6x^2y^2 - 4xy^3 - y^4\)

**Question 5:**

Expand \((5a + b)^5\).

A. \(295a^5 + 3125a^4b + 1250a^3b^2 + 250a^2b^3 + 25ab^4 + b^5\)
B. \(3125a^5 + 3125a^4b + 1250a^3b^2 + 250a^2b^3 + 25ab^4 + b^5\)
C. \(3125a^5 - 3125a^4b + 1250a^3b^2 - 250a^2b^3 + 25ab^4 - b^5\)
D. \(3125a^5 + 3125a^4b + 1250a^3b^2 + 250a^2b^3 + 25ab^4 + 10b^5\)

**Question 6:**

Expand \((x + 2y)^5\).

A. \(x^5 + 10x^4y + 40x^3y^2 + 80x^2y^3 + 80xy^4 + 32y^5\)
B. \(x^5 - 10x^4y + 40x^3y^2 - 80x^2y^3 + 80xy^4 - 32y^5\)
C. \(x^5 - 10x^4y + 40x^3y^2 - 80x^2y^3 + 80xy^4 - 32y^5\)
D. \(x^5 - 10x^4y + 40x^3y^2 - 80x^2y^3 + 80xy^4 - 32y^5\)

**Question 7:**

What is the sixth term of \((a - y)^7\)?

A. \(21a^3y^4\)
B. \(-21a^3y^4\)
C. \(21a^2y^5\)
D. \(-21a^2y^5\)

**Question 8:**

What is the 6th term of \((2x - 3y)^{11}\)?

A. \(7,187,024x^6y^5\)
B. \(-7,185,024x^6y^5\)
C. \(7,185,024x^6y^5\)
D. \(-7,187,024x^6y^5\)

**Question 9:**

What is the sixth term of \((x + y)^8\)?

A. \(56x^3y^5\)
B. \(36x^3y^5\)
C. \(36x^4y^4\)
D. \(56x^4y^4\)

**Question 10:**

What is the seventh term of \((x + 4)^8\)?

A. \(114,688x^2\)
B. \(114,688x^3\)
C. \(114,688x^4\)
D. \(114,688x^5\)

Answer :

QUESTION 1

We want to expand [tex](x-2)^6[/tex].


We apply the binomial theorem which is given by the formula

[tex](a+b)^n=^nC_0a^nb^0+^nC_1a^{n-1}b^1+^nC_2a^{n-2}b^2+...+^nC_na^{n-n}b^n[/tex].

By comparison,

[tex]a=x,b=-2,n=6[/tex].


We substitute all these values to obtain,


[tex](x-2)^6=^6C_0x^6(-2)^0+^6C_1x^{6-1}(-2)^1+^6C_2x^{6-2}(-2)^2+^6C_3x^{6-3}(-2)^3+^6C_4x^{6-4}(-2)^4+^6C_5x^{6-5}(-2)^5+^6C_6x^{6-6}(-2)^6[/tex].


We now simplify to obtain,

[tex](x-2)^6=^nC_0x^6(-2)^0+^6C_1x^{5}(-2)^1+^6C_2x^{4}(-2)^2+^6C_3x^{3}(-2)^3+^6C_4x^{2}(-2)^4+^6C_5x^{1}(-2)^5+^6C_6x^{0}(-2)^6[/tex].

This gives,

[tex](x-2)^6=x^6-12x^{5}+60x^{4}-160x^{3}(-2)^3+240x^{2}-1925x+64[/tex].


Ans:C

QUESTION 2


We want to expand

[tex](x+2y)^4[/tex].


We apply the binomial theorem to obtain,


[tex](x+2y)^4=^4C_0x^4(2y)^0+^4C_1x^{4-1}(2y)^1+^4C_2x^{4-2}(2y)^2+^4C_3x^{4-3}(2y)^3+^4C_4x^{4-4}(2y)^4[/tex].


We simplify to get,


[tex](x+2y)^4=x^4(2y)^0+4x^{3}(2y)^1+6x^{2}(2y)^2+4x^{1}(2y)^3+x^{0}(2y)^4[/tex].


We simplify further to obtain,


[tex](x+2y)^4=x^4+8x^{3}y+24x^{2}y^2+32x^{1}y^3+16y^4[/tex]


Ans:B


QUESTION 3

We want to find the number of terms in the binomial expansion,

[tex](a+b)^{20}[/tex].


In the above expression, [tex]n=20[/tex].


The number of terms in a binomial expression is [tex](n+1)=20+1=21[/tex].


Therefore there are 21 terms in the binomial expansion.


Ans:C


QUESTION 4


We want to expand

[tex](x-y)^4[/tex].


We apply the binomial theorem to obtain,


[tex](x-y)^4=^4C_0x^4(-y)^0+^4C_1x^{4-1}(-y)^1+^4C_2x^{4-2}(2y)^2+^4C_3x^{4-3}(-y)^3+^4C_4x^{4-4}(-y)^4[/tex].


We simplify to get,


[tex](x+2y)^4=^x^4(-y)^0+4x^{3}(-y)^1+6x^{2}(-y)^2+4x^{1}(-y)^3+x^{0}(-y)^4[/tex].


We simplify further to obtain,


[tex](x+2y)^4=x^4-4x^{3}y+6x^{2}y^2-4x^{1}y^3+y^4[/tex]


Ans: C


QUESTION 5

We want to expand [tex](5a+b)^5[/tex]


We apply the binomial theorem to obtain,

[tex](5a+b)^5=^5C_0(5a)^5(b)^0+^5C_1(5a)^{5-1}(b)^1+^5C_2(5a)^{5-2}(b)^2+^5C_3(5a)^{5-3}(b)^3+^5C_4(5a)^{5-4}(b)^4+^5C_5(5a)^{5-5}(b)^5[/tex].


We simplify to obtain,

[tex](5a+b)^5=^5C_0(5a)^5(b)^0+^5C_1(5a)^{4}(b)^1+^5C_2(5a)^{3}(b)^2+^5C_3(5a)^{2}(b)^3+^5C_4(5a)^{1}(b)^4+^5C_5(5a)^{0}(b)^5[/tex].


This finally gives us,


[tex](5a+b)^5=3125a^5+3125a^{4}b+1250a^{3}b^2+^250a^{2}(b)^3+25a(b)^4+b^5[/tex].


Ans:B

QUESTION 6

We want to expand [tex](x+2y)^5[/tex].

We apply the binomial theorem to obtain,

[tex](x+2y)^5=^5C_0(x)^5(2y)^0+^5C_1(x)^{5-1}(2y)^1+^5C_2(x)^{5-2}(2y)^2+^5C_3(x)^{5-3}(2y)^3+^5C_4(x)^{5-4}(2y)^4+^5C_5(x)^{5-5}(2y)^5[/tex].


We simplify to get,


[tex](x+2y)^5=^5C_0(x)^5(2y)^0+^5C_1(x)^{4}(2y)^1+^5C_2(x)^{3}(2y)^2+^5C_3(x)^{2}(2y)^3+^5C_4(x)^{1}(2y)^4+^5C_5(x)^{0}(2y)^5[/tex].


This will give us,

[tex](x+2y)^5=x^5+^10(x)^{4}y+40(x)^{3}y^2+80(x)^{2}y^3+80(x)y^4+32y^5[/tex].


Ans:A


QUESTION 7

We want to find the 6th term of [tex](a-y)^7[/tex].


The nth term is given by the formula,

[tex]T_{r+1}=^nC_ra^{n-r}b^r[/tex].

Where [tex]r=5,n=7,b=-y[/tex]


We substitute to obtain,


[tex]T_{5+1}=^7C_5a^{7-5}(-y)^5[/tex].


[tex]T_{6}=-21a^{2}y^5[/tex].


Ans:D


QUESTION 8.

We want to find the 6th term of [tex](2x-3y)^{11}[/tex]


The nth term is given by the formula,

[tex]T_{r+1}=^nC_ra^{n-r}b^r[/tex].

Where [tex]r=5,n=11,a=2x,b=-3y[/tex]


We substitute to obtain,


[tex]T_{5+1}=^{11}C_5(2x)^{11-5}(-3y)^5[/tex].


[tex]T_{6}=-7,185,024x^{6}y^5[/tex].


Ans:D

QUESTION 9

We want to find the 6th term of [tex](x+y)^8[/tex].


The nth term is given by the formula,

[tex]T_{r+1}=^nC_ra^{n-r}b^r[/tex].

Where [tex]r=5,n=8,a=x,b=y[/tex]


We substitute to obtain,


[tex]T_{5+1}=^8C_5(x)^{8-5}(y)^5[/tex].


[tex]T_{6}=56a^{3}y^5[/tex].


Ans: A


We want to find the 7th term of [tex](x+4)^8[/tex].


The nth term is given by the formula,

[tex]T_{r+1}=^nC_ra^{n-r}b^r[/tex].

Where [tex]r=6,n=8,a=x,b=4[/tex]


We substitute to obtain,


[tex]T_{6+1}=^8C_5(x)^{8-6}(4)^6[/tex].


[tex]T_{7}=114688x^{2}[/tex].


Ans:A





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