2 polynomial question!?

[匿名]

2016-07-07 2:26 am

1. The remainder of polynomial P(x) is 3 when divided by (x-2) and 2 when divided by (x-3). If the remainder P(x) is (px+q) when divided by (x^2 - 5x + 6), find the values of p and q.

2. The polynomial P(x) is given by P(x)= x^4-x^3-x^2+x.
(a) Use the factor theorem to show that (x-1) is factor of P(x)
(b) Write P(x) is the form of (x-1)(ax)(bx^2-c), giving he values of a, b and c.
(c) Hence, solve the equation P(x)=0
(d) Using your solutions to P(x) =0, write down the solutions of the equation P(x+1)=0.

All help is greatly appreciated ;-;

回答 (1)

Lopez

2016-07-07 9:40 am

✔ 最佳答案

1.
Let P(x) = A(x)(x-2) + 3 = B(x)(x-3) + 2 = C(x)( x^2 - 5x + 6 ) + px + q
P(2) = 3 = 2p + q ..... (1)
P(3) = 2 = 3p + q ..... (2)
(2) - (1) , we get p = - 1
By (1) , q = 3 - 2p = 3 - 2(-1) = 5
Ans: p = - 1 , q = 5

--------------------------------------------------------------------------------------------

2.
(a)
P(1) = 1 - 1 - 1 + 1 = 0
By the factor theorem, (x-1) is factor of P(x)

(b)
P(x)
= x^4 - x^3 - x^2 + x
= x ( x^3 - x^2 - x + 1 )
= x ( x - 1 )( x^2 - 1 ) , by the synthetic division
= ( x - 1 ) x ( x^2 - 1 )
Hence,
a = b = c = 1 ..... Ans

(c)
P(x)
= x ( x - 1 )( x^2 - 1 )
= x ( x - 1 )( x + 1 )( x - 1 )
= ( x + 1 ) x ( x - 1 )^2 = 0
x = -1 , 0 , 1 ( double roots ) ..... Ans

(d)
P(x+1) = ( x + 2 )( x + 1 ) x^2 = 0
x = -2 , -1 , 0 ( double roots ) ..... Ans

👍 0 👎 0