If P(x) is a polynomial and a real number such that P(a)=0. Then P(x) is a multiple of x - a, i.e. there exist

Question

If P(x) is a polynomial and a real number such that P(a)=0. Then P(x) is a multiple of x - a, i.e. there exists a polynomial Q(x) such that P(x) = Q(x) (x - a).

Answer 1

If P(x) were to be divided by (x-a), then the result would be

P(x)=Q(x)(x-a) + R ....1
where R is a constant
The above comes from Remainder theorem
The above is true for all x

so P(a) = R by putting x = a on both sides
but P(a) = 0 (given)
so R= 0
so P(x) = Q(x)(x-a)

This is incidentally known as factor theorem and is used to factorise polynomials.

a need not be real. This is true for any a.

Answer 2

Well, yeah, that would follow. The statement is saying the polynomial crosses the x-axis at (a,0), therefore (x - a) is a factor of the polynomial.

Now, coming up with a formal proof of this is another question entirely. :)

Answer 3

If P(x) were to be divided by (x-a), then the result would be

P(x)=Q(x)(x-a) + R(x)

where deg R(x) < deg (x-a), so deg R(x)=0. That is, R(x) is a constant b.

If b is zero, then (x-a) does divide P(x).

If b is nonzero, then (x-a) does not divide P(x), but then P(a) is nonzero, which is a contradiction.

Therefore (x-a) divides P(x).

Answer 4

yes
this is correct