State and prove remainder theorem and …

Remainder theorem: Let p(x) be any polynomial of degree greater than or equal to one and let a be any real number. If p(x) is divided by the linear polynomial x – a, then the remainder is p(a).

Proof : Let p(x) be any polynomial with degree greater than or equal to 1. Suppose

that when p(x) is divided by x – a, the quotient is q(x) and the remainder is r(x),

i.e.,p(x) = (x – a) q(x) + r(x)

Since the degree of x – a is 1 and the degree of r(x) is less than the degree of x – a,

the degree of r(x) = 0. This means that r(x) is a constant, say r.

So, for every value of x, r(x) = r.

Therefore, p(x) = (x – a) q(x) + r

In particular, if x = a, this equation gives us

p(a) = (a-a) q(a) + r

      = r

which proves the theorem.

Factor Theorem:  x – a is a factor of the polynomial p(x), if p(a) = 0. Also, if x – a is a factor of p(x), then p(a) = 0, where a is any real number. This is an extension to remainder theorem where remainder is 0, i.e. p(a) = 0.

Numerical: Examine whether x + 2 is a factor of p(x)= x³ + 3x² + 5x + 6

Solution : The zero of x + 2 is –2.  As, per factor theorem, x+2 is factor of p(x) if p(-2) = 0.

 p(–2) = (–2)³ + 3(–2)² + 5(–2) + 6 =  0.

Thus, x+2 is factor of p(x)= x³ + 3x² + 5x + 6