Question

if (y+1) and (y+2) are factors of the polynomial y3 + 3y2 - 3py + q, find p and q.

Answer 1

y+1=0 = y=(-1)
since it is factor of given equation
put y = -1
$({ - 1})^{3} + 3( { - 1}) ^{2} - 3p( - 1) + q$
-1+3+3p+q
2 + 3p+q..........1


now put y=-2
$( { - 2})^{3} + 3( { - 2})^{2} - 3p( - 2) + q$
-8+12+6p+q..........2

solve 1 and 2

Answer 2

Hi ,

Let p( y ) = y³ + 3y²-3py + q

It is given that ,

( y + 1 ) , ( y + 2 ) are the factors of

p( y ) ,

By , factor theorem ,

p( - 1 ) = 0 and p( - 2 ) = 0

( -1 )³ + 3( - 1 )² - 3p( -1 ) + q = 0

-1 + 3 + 3p + q = 0

3p + q +2 = 0

q = - 3p - 2 ----( 1 )

p( - 2 ) = 0

( -2 )³ + 3( -2 )² - 3p( -2 ) + q = 0

- 8 + 12 + 6p + q = 0

4 + 6p + q = 0

q = - 4 - 6p ---( 2 )

( 1 ) = ( 2 )

-3p - 2 = -4 - 6p

6p - 3p = - 4 + 2

3p = - 2

p = - 2/3

Put p = - 2/3 in equation ( 1 ) , we get

q = - 3( -2/3 ) - 2

= 2 - 2

q = 0

Therefore ,

p = -2/3 , q = 0

I hope this helps you.

: )