Question

match each binomial expression with the set of coefficient of the term obtain by expanding the expression

Answer 1

$(2x+y)^3$ = $8x^3+12x^2y+6xy^2+y^3$

The coefficients are  8 , 12, 6, 1

$(2x+3y)^4$= $16x^4+96x^3y+216x^2y^2+216xy^3+81y^4$

The coefficients are  16, 96, 216, 216, 81

$(3x+2y)^3$=$27x^3+54x^2y+36xy^2+8y^3$

The coefficients are  27, 54,36 , 8

$(x+y)^4$ = x^4+4x^3y+6x^2y^2+4xy^3+y^4

The coefficients are 1, 4, 6, 4, 1

Answer 2

Answer:

I   ...  3

II ...  4

III  ...  2

IV ... 1

Step-by-step explanation:

Recall binomial theorem as

(x+y)^n = x^n+nCr x^(n-1)Y+....+y^n

Using the above we find that

(x+y)^4 = x^4+4x^3+6x^2+4x+1

Hence 4 matches with 1.

Next option 27,54,36,8

are from (3x+2y)^3 because

(3x+2y)^3 = 27x^3+8y^3+18xy(3x+2y)

Option 2 for question 3.

Next is (2x+y)^3 will have 4 terms with coefficients as

8, 12,6,1

So I question for option 4.

The last question 2 = (2x+3y)^4

= (2x)^4+4(2x)^3(3y) + 6(2x)^2(3y)^2+4(2x)(3y)^3+(3y)^4

Hence coefficients are 16,96, 216,216 and 81

I   ...  3

II ...  4

III  ...  2

IV ... 1