Question

show that y-1 is a factor of yto the power 30-1 and
also of y to the power 31-1​

Answer 1

SOLUTION

TO PROVE

$\sf{1. \: \:( y - 1 ) \: \: is \: a \: factor \: of \: \: ({y}^{30} - 1 ) \: }$

$\sf{2. \: \:( y - 1 ) \: \: is \: a \: factor \: of \: \: ({y}^{31} - 1 ) \: }$

EVALUATION

Let f(y)= y - 1

For Zero of the polynomial f(y) we have

$\sf{f(y) = 0}$

$\implies \sf{y - 1 = 0}$

$\implies \sf{y = 1}$

So the Zero of the polynomial f(y) is 1

ANSWER TO QUESTION : 1

$\sf{Let \: \: g(y) = {y}^{30} - 1 }$

Now by the Remainder Theorem the Remainder when g(y) is divided by f(y) is

$= \sf{g(1)}$

$= \sf{ {(1)}^{30} - 1}$

$= \sf{1 - 1 \: }$

$\sf{ = 0}$

Since the Remainder = 0

$\sf{ \therefore \: \:( y - 1 ) \: \: is \: a \: factor \: of \: \: ({y}^{30} - 1 ) \: }$

ANSWER TO QUESTION : 2

$\sf{Let \: \: h(y) = {y}^{31} - 1 }$

Now by the Remainder Theorem the Remainder when h(y) is divided by f(y) is

$= \sf{h(1) \: }$

$\sf{ = {(1)}^{31} - 1 \: }$

$\sf{ = 1 - 1 \: }$

$\sf{ = 0}$

Since the Remainder = 0

$\sf{ \therefore \: \:( y - 1 ) \: \: is \: a \: factor \: of \: \: ({y}^{31} - 1 ) \: }$

Hence proved

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