1. ## Polynomial

Polynomials p(x) and q(x) are given by relationship q(x)=p(x)(x^5-2x+2).

a) If x-2 is a factor of p(x)-5 , find the remainder when q(x) is divided by x-2.

b) If p(x) is of the form x^2+ax+b and x-1 is a factor of p(x)-5, find the values of a and b.

Can you help me?

2. Originally Posted by MichaelLight
Polynomials p(x) and q(x) are given by relationship q(x)=p(x)(x^5-2x+2).

a) If x-2 is a factor of p(x)-5 , find the remainder when q(x) is divided by x-2.

b) If p(x) is of the form x^2+ax+b and x-1 is a factor of p(x)-5, find the values of a and b.Can you help me?
solution to first part. x-2 divides p(x)-5 so p(2)-5=0
you need to find the remainder when q(x) is divided by x-2 so you need to find the value of q(2). now q(2)=p(2)(2^5-2*2+2). p(2)=5. just plug in the values.

3. the second part.
if the only data given is that x-1 is a factor of p(x)-5 this can lead to only one equation and that is p(1)=5, that is, 1+a+b=5.
this wont give you the values of a and b. probably the data given in the first part of the question should also be used, that is x-2 is also a factor of p(x)-5. this gives another equation in a and b and that is p(2)=5 so 4+2a+b=5. solve these two simultaneous linear equations.

4. Originally Posted by MichaelLight
Polynomials p(x) and q(x) are given by relationship q(x)=p(x)(x^5-2x+2).

a) If x-2 is a factor of p(x)-5 , find the remainder when q(x) is divided by x-2.
The remaind erof q(x) when divided by x- 2 is q(2)
q(x)= p(x)(x^5-2x+ 2)+ 5(x^5- 2x+ 2)- 5(x^5- 2x+ 2)= (p(x)- 5)(x^5- 2x+ 2)+ 5(x^5- 2x+ 2)
Since x- 2 is a factor of p(x)- 5, p(2)- 5= 0. What is q(2)?

b) If p(x) is of the form x^2+ax+b and x-1 is a factor of p(x)-5, find the values of a and b.

Can you help me?
Are we to assume that this is the same p(x) as in (a)? If so p(x)- 5 also has a factor of x- 2 and so x^2+ ax+ b- 5= (x- 2)(x- 1). Multiply on the right, add 5 to both sides, and compare coefficients.

If not, the best we can do is: p(x)= x^2+ ax+ b so p(x)- 5= x^2+ ax+ b-5. Since x- 1 is a factor of that 1+ a+ b- 5= 0 and there are an infinite number of solutions.