Question

1) For P(x) = x3 – 2 , P(5) = ___________. A. 125 B. 123 C. 120 D. 75 2) The zero of 3x – 5 is _________. A. 5/3 B. – 5/3 C. 3/5 D. – 3/ 5 3) The sum of the zeroes of a cubic polynomial is __________. A. – b/a B. c/a C. – d/a D. – c/a​

Answer 1

Answer:

1) For P(x) = x3 – 2 , P(5) = ___________. A. 125 B. 123 C. 120 D. 75 2) The zero of 3x – 5 is _________. A. 5/3 B. – 5/3 C. 3/5 D. – 3/ 5 3) The sum of the zeroes of a cubic polynomial is __________. A. – b/a B. c/a C. – d/a D. – c/a

Answer 2

Answer:

Coefficient of x⁶y³ is 672.

Step-by-step explanation:

General term of expansion (a+b)ⁿ is

\bf \: T_{r+1} = \: \: ^nC_r \: \: \large \frak{ a ^{n−r} b ^r}T

r+1

=

n

C

r

a

n−r

b

r

For (x+2y)⁹,

Putting n =9, a=x, b=2y

\begin{gathered} \bf \: T_{r+1} = \: \: ^{9} C_r (x) ^{9−r} (2y) ^r \\ \\ \bf \: T _{r+1} = \: \: ^{9} C_r (x) ^{9−r} .(y) ^r .(2) ^r\end{gathered}

T

r+1

=

9

C

r

(x)

9−r

(2y)

r

T

r+1

=

9

C

r

(x)

9−r

.(y)

r

.(2)

r

Comparing with x⁶ y³ , we get, r = 3

Therefore,

\begin{gathered} \bf \: T _{r+1} \\ \: \tt ^9C_3 (x)^9−3 .y³ .2³ \\ \tt\: \: 9! (2)³× x⁶ × y³) / (3!.6!) \\ \: \tt 672x⁶ y³\end{gathered}

T

r+1

9

C

3

(x)

9

−3.y³.2³

9!(2)³×x⁶×y³)/(3!.6!)

672x⁶y³