Remainder Theorem
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).
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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.
- Remainder obtained on dividing polynomial p(x) by x – a is equal to p(a) .
- If a polynomial p(x) is divided by (x + a) the remainder is the value of p(x) at x = –a.
- (x – a) is a factor of polynomial p(x) if p(a) = 0
- (x + a) is a factor of polynomial p(x) if p(–a) = 0
- (x – a) (x – b) is a factor of polynomial p(x), if p(a) = 0 and p(b) = 0.
- If a polynomial p(x) is divided by (ax – b), the remainder is the value of p(x) at x = b/a
- If a polynomial p(x) is divided by (b – ax), the remainder is equal to the value of p(x) at x = b/a.
- (ax – b) is a factor of polynomial p(x) if p(b/a) = 0.
Remainder Theorem Example Problems With Solutions
Example 1: Find the remainder when 4x3 – 3x2 + 2x – 4 is divided by x – 1
Solution: Let p(x) = 4x3 – 3x2 + 2x – 4
When p(x) is divided by (x – 1), then by remainder theorem, the required remainder will be p(1)
p(1) = 4 (1)3 – 3(1)2 + 2(1) – 4
= 4 × 1 – 3 × 1 + 2 × 1 – 4
= 4 – 3 + 2 – 4 = – 1
Example 2: Find the remainder when 4x3 – 3x2 + 2x – 4 is divided by x + 1
Solution: Let p(x) = 4x3 – 3x2 + 2x – 4
When p(x) is divided by (x + 2), then by remainder theorem, the required remainder will be p (–2).
p(–2) = 4 (–2)3 – 3 (–2)2 + 2(–2) – 4
= 4 × (–8) – 3 × 4 – 4 – 4
= – 32 – 12 – 8 = – 52
Example 3: Find the remainder when 4x3 – 3x2 + 2x – 4 is divided by \(\text{x + }\frac{1}{2}\)
Solution: Let p(x) = 4x3 – 3x2 + 2x – 4
When p(x) is divided by \(\text{x + }\frac{1}{2}\), then by remainder theorem, the required remainder will be \(p\left( -\frac{1}{2} \right)={{\left( -\frac{1}{2} \right)}^{3}}-3{{\left( -\frac{1}{2} \right)}^{2}}+2\left( -\frac{1}{2} \right)-4\)
\(=4\times \left( -\frac{1}{8} \right)-3\times \frac{1}{4}-2\times \frac{1}{2}-4\)
\(=-\frac{1}{2}-\frac{3}{4}-1-4=\frac{-2-3-20}{4}\)
\(=-\frac{25}{4}\)
Example 4: Determine the remainder when the polynomial p(x) = x4 – 3x2 + 2x + 1 is divided by x – 1.
Solution: By remainder theorem, the required remainder is equal to p(1).
Now, p (x) = x4 – 3x2 + 2x + 1
p(1) = (1)4 – 3×12 + 2 × 1 + 1
= 1 – 3 + 2 + 1 = 1
Hence required remainder = p(1) = 1
Example 5: Find the remainder when the polynomial f(x) = 2x4 – 6x3+ 2x2 – x + 2 is divided by x + 2.
Solution: We have, x + 2 = x – (–2). So, by remainder theorem, when f(x) is divided by (x–(–2)) the remainder is equal to f(–2).
Now, f(x) = 2x4 – 6x3+ 2x2 – x + 2
⇒ f(–2) = 2 (–2)4 – 6(–2)3 + 2(–2)2 – (–2)+2
⇒ f(–2) = 2×16 – 6 × –8 + 2 × 4 + 2 + 2
⇒ f(–2) = 32 + 48 + 8 + 2 + 2 = 92
Hence, required remainder = 92
Example 6: Find the remainder when p(x) = 4x3 – 12x2 + 14x – 3 is divided by g(x) = x – 1/2
Solution:
Example 7: If the polynomials ax3 + 4x2 + 3x – 4 and x3– 4x + a leave the same remainder when divided by (x–3), find the value of a.
Solution: Let p(x) = ax3 + 4x2 + 3x – 4 and
q(x) = x3 – 4x + a be the given polynomials. The remainders when p(x) and q(x) are divided by (x–3) are p(3) and q(3) respectively.
By the given condition, we have
p(3) = q(3)
Example 8: Let R1 and R2 are the remainders when the polynomials x3 + 2x2 –5ax–7 and x3 + ax2 – 12x + 6 are divided by x + 1 and x – 2 respectively. If 2R1 + R2 = 6, find the value of a.
Solution: Let p(x) = x3 + 2x2 –5ax–7 and
