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Proof of remainder theorem:
Let q(x) be the quotient and r(x) be the remainder obtained when the polynomial p(x) is divided by (x–a).
Then, p(x) = (x – a) q(x) + r(x), where r(x) = 0 or some constant.
Let r(x) = c, where c is some constant. Then
p(x) = (x–a) q(x) + c
Putting x = a in p(x) = (x–a) q(x) + c, we get
p(a) = (a–a) q(a) + c ⇒ p(a) = 0 x q(a) + c ⇒ p(a) = c
This shows that the remainder is p(a) when p(x) is divided by (x–a).
Proof of this factor theorem
Let p(x) be a polynomial of degree greater than or equal to one and a be a real number such that p(a) = 0. Then, we have to show that (x – a) is a factor of p(x).
Let q(x) be the quotient when p(x) is divided by (x – a).
By remainder theorem : Dividend = Divisor x Quotient + Remainder
p(x) = (x – a) x q(x) + p(a) [Remainder theorem]
⇒ p(x) = (x – a) x q(x) [p(a) = 0]
⇒ (x – a) is a factor of p(x)
Conversely, let (x – a) be a factor of p(x). Then we have to prove that p(a) = 0
Now, (x – a) is a factor of p(x)
⇒ p(x), when divided by (x – a) gives remainder zero.
But, by the remainder theorem, p(x) when divided by (x – a) gives the remainder equal to p(a). ∴ p(a) = 0
Let q(x) be the quotient and r(x) be the remainder obtained when the polynomial p(x) is divided by (x–a).
Then, p(x) = (x – a) q(x) + r(x), where r(x) = 0 or some constant.
Let r(x) = c, where c is some constant. Then
p(x) = (x–a) q(x) + c
Putting x = a in p(x) = (x–a) q(x) + c, we get
p(a) = (a–a) q(a) + c ⇒ p(a) = 0 x q(a) + c ⇒ p(a) = c
This shows that the remainder is p(a) when p(x) is divided by (x–a).
Proof of this factor theorem
Let p(x) be a polynomial of degree greater than or equal to one and a be a real number such that p(a) = 0. Then, we have to show that (x – a) is a factor of p(x).
Let q(x) be the quotient when p(x) is divided by (x – a).
By remainder theorem : Dividend = Divisor x Quotient + Remainder
p(x) = (x – a) x q(x) + p(a) [Remainder theorem]
⇒ p(x) = (x – a) x q(x) [p(a) = 0]
⇒ (x – a) is a factor of p(x)
Conversely, let (x – a) be a factor of p(x). Then we have to prove that p(a) = 0
Now, (x – a) is a factor of p(x)
⇒ p(x), when divided by (x – a) gives remainder zero.
But, by the remainder theorem, p(x) when divided by (x – a) gives the remainder equal to p(a). ∴ p(a) = 0
1. If (x  1) is a factor of the polynomial p(x) = 3x^4  4x^3  ax +2 then find the value of ‘a’ ?
2. What are the possible expressions for the dimensions of a cuboid whose volume is given below ?
Volume = 12ky^2 + ky  20k.
3. If x = 2y + 6 then find the value of x^3  8y^3  36xy  216.
4. Prove that : (a^2 b^2)^3 + (b^2  c^2)^3 + (c^2  a^2)^3 = 3 (a + b) (b + c) (c + a) (a  b) (b  c) (c  a)
5. If remainder is same when polynomial p(x) = x3+ 8x2 + 17x + ax is divided by (x + 2) and (x + 1), find the value of a.
6. Factorize : x^3  3x^2  9x 5.
7. If ax^3 + bx^2 +x  6 has (x + 2) as a factor and leaves remainder 4 when divided by (x  2) , find the values of a and b.
8. If a + b + c = 6, find the value of : (2  a)^3 + (2  b)^3 + (2  c)^3  3(2  a)(2  b)(2  c).
2. What are the possible expressions for the dimensions of a cuboid whose volume is given below ?
Volume = 12ky^2 + ky  20k.
3. If x = 2y + 6 then find the value of x^3  8y^3  36xy  216.
4. Prove that : (a^2 b^2)^3 + (b^2  c^2)^3 + (c^2  a^2)^3 = 3 (a + b) (b + c) (c + a) (a  b) (b  c) (c  a)
5. If remainder is same when polynomial p(x) = x3+ 8x2 + 17x + ax is divided by (x + 2) and (x + 1), find the value of a.
6. Factorize : x^3  3x^2  9x 5.
7. If ax^3 + bx^2 +x  6 has (x + 2) as a factor and leaves remainder 4 when divided by (x  2) , find the values of a and b.
8. If a + b + c = 6, find the value of : (2  a)^3 + (2  b)^3 + (2  c)^3  3(2  a)(2  b)(2  c).
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