Question

find the factor of y cube + y square + y + 1 .

Answer 1

we know the factor theorem :

Let p( x ) be a polynomial of degree one or

more than 1 and a is a real number. Then ,

i ) x - a , will be a factor of p( x ) if p( a ) = 0

ii ) If x - a is a factor of p( x ) , then p(a ) = 0.

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Now ,

let us assume ,

p ( y ) = y² + 2y - 15 ,

g ( y ) = y³ + a ;

according to the problem given ,

( y + k ) is a factor of p ( y ) and g ( y ) .

1 ) By factor theorem ,

p ( - k ) = 0 ---( 1 )

g ( - k ) = 0 ---( 2 )

p ( - k ) = ( - k )² + 2( - k ) - 15 = 0

k² - 2k - 15 = 0

splitting the middle term ,

k² - 5k + 3k - 15 = 0

k ( k - 5 ) + 3 ( k - 5 ) = 0

( k - 5 ) ( k + 3 ) = 0

Therefore ,

k - 5 = 0 or k + 3 = 0

k = 5 or k = - 3

_________________________

g ( - k ) = 0

( - k )³ + a = 0

- k³ + a = 0

a = - k³---( 3 )

now substitute k = 5 and k = - 3 in

equation ( 3 ) we get

a = - 5³ = - 125

a = - ( - 3 )³ = - ( - 27 ) = 27

Therefore ,

if k = 5 then a = - 125

or

if k = - 3 then a = 27

I hope this helps you.

: )

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Answer 2

$y^3+y^2+y+1 \\ \\ y^3+y+y^2+1 \\ \\ y(y^2+1)+1(y^2+1) \\ \\ \bold{(y+1)(y^2+1)}$

$(y+1)(y^2+1) \\ \\ (y+1)(y^2-(-1)) \\ \\ (y+1)(y^2-\iota^2) \\ \\ \bold{(y+1)(y+\iota)(y-\iota)}$

$Here, \ \ \iota=\sqrt{-1}$

$So that \ \ \iota^2=-1$

$Hope this helps. \\ \\ \\ Thank you. :-)$