**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 4x^{3} – 3x^{2} + 2x – 4 is divided by x – 1

**Solution: **Let p(x) = 4x^{3} – 3x^{2} + 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 4x^{3} – 3x^{2} + 2x – 4 is divided by x + 1

**Solution:** Let p(x) = 4x^{3} – 3x^{2} + 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 4x^{3} – 3x^{2} + 2x – 4 is divided by \(\text{x + }\frac{1}{2}\)

**Solution:** Let p(x) = 4x^{3} – 3x^{2} + 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) = x^{4} – 3x^{2} + 2x + 1 is divided by x – 1.

**Solution:** By remainder theorem, the required remainder is equal to p(1).

Now, p (x) = x^{4} – 3x^{2} + 2x + 1

p(1) = (1)^{4} – 3×1^{2} + 2 × 1 + 1

= 1 – 3 + 2 + 1 = 1

Hence required remainder = p(1) = 1

**Example 5: **Find the remainder when the polynomial f(x) = 2x^{4} – 6x^{3}+ 2x^{2} – 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) = 2x^{4} – 6x^{3}+ 2x^{2} – 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) = 4x^{3} – 12x^{2} + 14x – 3 is divided by g(x) = x – 1/2

**Solution: **

**Example 7: **If the polynomials ax^{3} + 4x^{2} + 3x – 4 and x^{3}– 4x + a leave the same remainder when divided by (x–3), find the value of a.

**Solution: **Let p(x) = ax^{3} + 4x^{2} + 3x – 4 and

q(x) = x^{3 }– 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 R_{1} and R_{2} are the remainders when the polynomials x^{3} + 2x^{2} –5ax–7 and x^{3} + ax^{2} – 12x + 6 are divided by x + 1 and x – 2 respectively. If 2R_{1} + R_{2} = 6, find the value of a.

**Solution: **Let p(x) = x^{3} + 2x^{2} –5ax–7 and