2p3 – xp2 –yp-14 has( p+2) as a factor and leaves a remainder - 26 when divided by ( p+3),find x and y.
Answers
Given :-
2p³ - xp² - yp -14 has (p+2) as a factor.
It leaves a remainder -26 when divided by (p+3).
To find :-
The values of x and y .
Solution :-
Given cubic polynomial is
P(p) = 2p³-xp²-yp-14
Given factor of it = p+2
We know that By Factor Theorem,
By Factor Theorem,P(-2) = 0
=> 2(-2)³-x(-2)²-y(-2)-14 = 0
=> 2(-8)-x(4)-y(-2)-14 = 0
=> -16 -4x+2y-14 = 0
=> -4x+2y -30 = 0
=> -2(2x+y+15) = 0
=> 2x+y+15 = 0/-2
=> 2x+y+15 = 0
=> 2x+y = -15 --------------(1)
Given that
P(p) leaves the remainder -26 when divided by (p+3)
We know that By Remainder Theorem
By Remainder TheoremP(-3) = -26
=> 2(-3)³-x(-3)²-y(-3)-14 = -26
=> 2(-27)-(9)x+3y-14 = -26
=> -54-9x+3y-14 = -26
=> -9x+3y -68 = -26
=> -9x+3y = -26+68
=> -9x+3y = 42
=> -3(3x-y) = 42
=> 3x-y = 42/-3
=> 3x-y = -14 -------------(2)
On adding (1)&(2)
2x+y = -15
3x-y = -14
(+)
__________
5x+0 = -29
___________
x = -29/5
From (1)
2(-29/5)+y = -15
=> (-58/5) + y = -15
=> y = -15+(58/5)
=> y = (-75+58)/5
=> y = -17/5
Therefore, x = -29/5 and y = -17/5
Answer :-
The values of x and y are -29/5 and -17/5 respectively.
Used formulae:-
Remainder Theorem:-
Let P(x) be a polynomial of the degree greater than or equal to 1 and x-a is another linear polynomial if P (x) is divided by x-a then the remainder is P (a).
Factor Theorem :-
Let P(x) be a polynomial of the degree greater than or equal to 1 and x-a is another linear polynomial,if (x-a) is a factor of P(x) the. P(a) = 0 vice versa.