Today our topic of discussion is Remainder Theorem.
Remainder Theorem
Remainder Theorem
In the following example, if 6x² – 7x + 5 is divided by x – 1 ,then what is quotient and remainder?
(x-1) 6x² – 7x + (6x-1 (6x² – 6x)/(- x + 5) (- x + 1)/4
Here, 1 is divisor, 6x ^ 2 – 7x + 5 is dividen 0.6x – 1 is quotient and 4 is remainder.
We know, dividend = divisor x quotient + remainder
Now, if we indicate the dividend by f(x) the quotient by h(x) the remainder
r and the divisor by (x – a) , from the above formula, we get,
f(x) = (x – a) * h(x) + r this formula is true to all values of a. Substituting x = a in both sides we get,
f(a) = (a – a) * h(a) + r = 0h(a) + r = r
Hence, r = f(a)
Therefore, if f(x) is divided by (x – a) , the remainder is f(a) This formula is known as remainder theorem. That is, the remainder theorem gives the remainder when a polynomial f(x) of positive degree is divided by * (x – a) without performing actual division. In the above example if a = 1 , then f(x) = 6x² – 7x + 5 .
f(1) = 6 – 7 + 5 = 4 which is equal to the remainder. The degree of the divisor polynomial (-a) is 1, If the divisor is a factor of the dividend, the remainder will be zero and if it is not zero, the remainder will be a number other than zero.
Corollary 11. (x – a) will be a factor of f(x) if and only if f(a) = 0 .
Proof: Let, f(a) = 0 Therefore, according to remainder theorem, if f(x) is divided by (x – a) the remainder will be zero. Hence, (x – a) will be a factor of f(x)
Conversely, let, (x – a) is a factor of f(x)
Therefore, f(x) = (x – a) * h(x) where h(x) is a polynomial.
Putting x = a in both sides we get,
f(a) = (a – a) * h(a) = 0
therefore f(a) = 0
Hence, any polynomial f(x) will be divisible by (x – a) if and only if f(a) = 0 This formula is known as factor theorem.
Proposition 12. If degree of f(x) is positive and a ne0 then if f(x) is divided by (ax + b) the remainder will be f(- b/a)
Proof: Degree of divisor ax + b ( a ne0) is 1.
Hence, we can write: f(x) = (ax + b) * h(x) + r = a(x + b/a) * h(x) + r
f(x) = (x + b/a) * ah(x) + r
It is clear that, if f(x) is divided by the remainder will be r. (x + b/a) the quotient will be ah(x) and
Here, divisor = x – (- b/a)
Hence, according to remainder theorem
r = f(- b/a)
Therefore, if f(x) is divided by (ax + b) , the remainder will be (- b/a)
Corollary 13. If a ne0 the expression ax + b will be a factor of any polynomial
f(x) if and only if f (- dot b a )=0.
Proof: a ne0,ax+b=a(x+ b a ) = x – (- b/a) will be a factor of f(x) if and only if is a factor of f(x) . Hence, if and only if (x + b/a) f(- b/a) = 0
This method of determining the factor by using remainder theorem is called Vanishing method.
Example 30. Resolve into factors: x³ – x – 6
Solution: Here, f(x) = x³ – x – 6 is a polynomial. The factors of the constant -6 are pm1, pm2, plus/minus 3 , pm6.
Now, putting – 1,-1 we see that the value of f(x) is not zero. of f(x) is not zero.
But, putting x = 2 we see that the value Hence, f(2) = 2 ³ – 2 – 6 = 8 – 2 – 6 = 0
Therefore, x – 2 is a factor of f(x)
f(x) = x ³ – x – 6
= x ³ – 2x² + 2x² – 4x + 3x – 6
= x² * (x – 2) + 2x(x – 2) + 3(x – 2)
= (x – 2)(x² + 2x + 3)
Example 31.Resolve into factors: x³ – 3x * y² + 2y³ and x² + xy – 2y²
Solution: Here, consider r a variable and y a constant.
We consider the given expression a polynomial of r.
Let, f(x) = x ³ – 3x * y² + 2y³
Then, f(y) = y ³- 3y * y ² + 2y ³= 3y³ – 3y ³ = 0 .. (x – y) is a factor of f(x)
Now, 2-3xy²+2y³
= x³ – x² * y + x² * y – x * y² – 2x * y² + 2y³= x ² * (x – y) + xy(x – y) – 2y ² * (x – y)
= (x – y)(x ² + xy – 2y ²)
Again let, g(x) = x² + xy – 2y²
g(y) = y ² + y² – 2y² = 0
.. (x – y) is a factor of g(x)
… g(x) = x² + xy – 2y²
= x ² – xy + 2xy – 2y²
= x(x – y) + 2y(x – y)
= (x – y)(x + 2y)
x³ – 3x * y² + 2y ³ = (x – y)² * (x + 2y)
Example 32. Resolve into factors: 54x ^ 4 + 27x³* a – 16x – 8a
Solution: Let, f(x) = 54x ^ 4 + 27x³ * a – 16x – 8a
3 Then, (-10)-3(-1)+27 (0) – 16(-1)-80 1 f =54
= 27/8 * a ^ 4 – 27/8 * a ^ 4 + 8a – 8a = 0
x – (- 1/2 * a) = x + a/2 = 1/2 * (2x + a) is a factor of f(x)
Hence, (2x + a) is a factor of f(x)
Now, 54x ^ 4 + 27x³ * a – 16x – 8a
= 27x³ * (2x + a) – 8(2x + a)
= (2x + a)(27x ² – 8)
= (2x + a){(3x)³ – (2)³ }
= (2x + a)(3x – 2)(9x² + 6x + 4)
Example 33. g(a) = a³ + a² + 10a – 8 f(a) = a³ – 9 + (a + 1)³
1) If g(a) is divided by (a – 2) , then determine the remainder.
2) Resolve into factors: f(a)
Solution: 1) Given, g(a) = a³ + a² + 10a – 8 According to remainder theorem, if g(a) is divided by (a – 2) then the remainder will be g(2) .
.. y(2) = 2³ + 2² + 10 * 2 – 8 = 8 + 4 + 20 – 8 = 32 – 8 = 24
.. g(2) = 24
Calculated remainder is 24.
f(a) = a³ – 9 + (a + 1)³
f(a) is a polynomial, if we put a = 1 then the result of the polynomial will be zero.
Therefore (a – 1) is a factor of the polynomial.
f(a) = a³ – 9 + a³ + 3a² + 3a + 1 = 2a³ + 3a² + 3a – 8
= 2a ³ – 2a² + 5a ² – 5a + 8a – 8
= 2a² * (a – 1) + 5a(a – 1) + 8(a – 1) – (a – 1) * (2a² + 5a + 8)
∴ a³ – 9 + (a + 1)³ = (a – 1)(2a² + 5a + 8)
Work: Resolve into factors:
1) x ³ – 21x – 20
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