Answer:
See solution
Step-by-step explanation:
Use formula for the binomial expansion:
[tex](a+b)^n=C_n^0a^n+C_n^1a^{n-1}b+C_n^2a^{n-2}b^2+\dots+C_n^{n-1}ab^{n-1}+C_n^nb^n.[/tex]
Note that [tex]C_n^0=C_n^n=1,\ C_n^1=C_n^{n-1}=n.[/tex]
1. For the expression [tex](x+2)^6:[/tex]
[tex](x+2)^6=x^6+C_6^1x^5\cdot 2+C_6^2x^4\cdot 2^2+C_6^3x^3\cdot 2^3+C_6^4x^2\cdot 2^4+C_6^5x\cdot 2^5+2^6=x^6+6\cdot 2x^5+15\cdot 4x^4+20\cdot 8x^3+15\cdot 16x^2+6\cdot 32x+64=x^6+12x^5+60x^4+160x^3+240x^2+192x+64.[/tex]
2. For the expression [tex](x-4)^4:[/tex]
[tex](x-4)^4=x^4-C_4^1x^3\cdot 4+C_4^2x^24^2-C_4^3x\cdot 4^3+4^4=x^4-4x^3\cdot 4+6x^2\cdot 16-4x\cdot 64+256=x^4-16x^3+96x^2-256x+256.[/tex]
3. For the expression [tex](2x+3)^5:[/tex]
[tex](2x+3)^5=(2x)^5+C_5^1(2x)^4\cdot 3+C_5^2(2x)^3\cdot 3^2+C_5^3(2x)^2\cdot 3^3+C_5^4(2x)\cdot 3^4+3^5=32x^5+5\cdot 16x^4\cdot 3+10\cdot 8x^3\cdot 9+10\cdot 4x^2\cdot 27+5\cdot 2x\cdot 81+243=32x^5+240x^4+720x^3+1080x^2+810x+243.[/tex]
4. For the xpression [tex](2x-3y)^4:[/tex]
[tex](2x-3y)^4=(2x)^4-C_4^1(2x)^3\cdot (3y)+C_4^2(2x)^2\cdot (3y)^2-C_4^3(2x)\cdot (3y)^3+(3y)^4=16x^4-4\cdot 8x^3\cdot 3y+6\cdot 4x^2\cdot 9y^2-4\cdot 2x\cdot 27y+81y^4=16x^4-96x^3y+216x^2y^2-216xy^3+81y^4.[/tex]
5. In the expansion of [tex](3a + 4b)^8[/tex] each term has the degree of 8. Since product [tex]a^2b^3[/tex] has degree 2+3=5, this term is not possible. Since product [tex]ab^8[/tex] has degree 1+8=9, this term is not possible. Similarly, term [tex]a^6b^5[/tex] has degree 6+5=11 and is not a term of expansion [tex](3a + 4b)^8.[/tex] Products [tex]a^5b^3,\ b^8,\ a^4b^4,\ a^8,\ ab^7[/tex] have degree 8 (5+3=8=4+4=8=1+7), these products are possible terms with corresponding coefficients.
