Factors of Polynomial –
A polynomial g(x) is called a factor of the polynomial f(x), if g(x) divides f(x) exactly giving 0 as reminder. Factor Theorem –
Let f(x) be a given polynomial and α be a given real number. Then,
(x – α) is a factor of f(x) => f(α) = 0
Proof. Let f(x) be a given polynomial and α be a given real number
As we know by Reminder Theorem that when f(x) is divided by (x – α), then reminder = f(α)
Now, (x – α) is a factor of f(x) => Reminder = 0
=> f(α) = 0
Hence, (x – α) is a factor of f(x) => f(α) = 0
Deduction 1. Prove that (x + α) is a factor of f(x) => f(- α) = 0
Proof.– We may write, (x + α) = [x – (- α)]
So, (x + α) is a factor of f(x) => f(- α) = 0
- b
Deduction 2. Prove that (ax + b) is a factor of f(x) => f(-----) = 0
a
b
Proof.– We may write, (ax + b) = a (x + ------)
a
- b
= a [x – (-----)]
a
- b
so, (ax + b) is a factor of f(x) => f (-----)
a
Example.1) Show that (x – 1) is a factor of (x³- 7x²+ 14x – 8). Hence, completely factorize the expression.
Ans.) Let, f(x) = x³- 7x²+ 14x – 8
We know that (x – 1) is a factor of f(x) => f(1) = 0
Now, f(1) = x³- 7x²+ 14x – 8
= 1³- (7 X 1³) + (14 X 1) – 8
= 1 – 7 + 14 – 8
= 15 – 15
= 0
So, it is proved that (x – 1) is a factor of f(x)
On dividing f(x) = x³- 7x²+ 14x – 8 by (x – 1),
x – 1 ) x³ - 7x² + 14x – 8 ( x² + 6x + 8
x³ - x²
----------------
- 6x²+ 14x
6x²- 6x
----------------
8x – 8
8x – 8
----------
0
We get, (x²- 6x + 8) as quotient.
So, f(x) = x³- 7x²+ 14x – 8
= (x – 1) (x²- 6x + 8)
= (x – 1) (x²- 4x - 2x + 8)
= (x – 1) [x (x - 4) - 2 (x - 4)]
= (x – 1) (x - 4) (x - 2)
Hence, x³- 7x²+ 14x – 8 = (x – 1) (x - 4) (x - 2) (Ans.)
Example 2) Given that (x + 2) and (x + 3) are factors of 2x³+ ax²+ 7x – b. Determine the value of a & b.
Ans.) Let f(x) = 2x³+ ax²+ 7x – b
Since, (x + 2) is a factor of f(x), we must have f(-2) = 0
Again since (x + 3) is a factor of f(x), we must have f(-3) = 0
Now, f(-2) = 0 => [{2 X (-2)³} + {a X (-2)²} + {7 X (-2)} – b] = 0
=> (- 16 + 4a – 14 – b) = 0
=> 4a – b = 30 …………………….(i)
And, f(-3) = 0 => [{2 X (-3)³} + {a X (-3)²} + {7 X (-3)} – b] = 0
=> (- 54 + 9a – 21 – b) = 0
=> 9a – b = 75 ………………. (ii)
On subtracting (i) from (ii), we get –
9a – b = 75
4a – b = 30
---------------
5a = 45
=> a = 9
Now we will put the value of a = 9 in (i), and we get –
4a – b = 30 …………………….(i)
=> (4 X 9) – b = 30
=> 36 – 30 = b
=> b = 6
Hence, a = 9, and b = 6 (Ans.)
Example.3) If (x – 2) is a factor of the expression 2x³+ ax²+ bx – 14 and when the expression is divided by (x – 3), it leaves a reminder 52. Find the volume of a & b.
Ans.) Let f(x) = 2x³+ ax²+ bx – 14
Since (x – 2) is a factor of f(x), we must have f(2) = 0
Now, f(2) = 0 => [(2 X 2³) + (a X 2²) + (b X 2) – 14 = 0
=> (16 + 4a + 2b – 14) = 0
=> 4a + 2b + 2 = 0
=> 2a + b + 1 = 0
=> 2a + b = - 1 ……………..(i)
Again, by Reminder Theorem, on dividing f(x) by (x – 3), we get f(3) as reminder
So, f(3) = 52 => [(2 X 3³) + (a X 3²) + (b X 3) – 14 = 52
=> (54 + 9a + 3b – 14) = 52
=> 9a + 3b + 40 = 52
=> 9a + 3b = 52 – 40 = 12
=> 3a + b = 4 ………………….(ii)
On subtracting (i) from (ii), and we get –
3a + b = 4
2a + b = -1
----------------
a = 5
now, we will substitute the value of a = 5 in (i), and we get –
2a + b = - 1 .................. (i)
=> (2 X 5) + b = - 1
=> 10 + b = - 1
=> b = - 11
Hence a = 5, and b = -11 (Ans.)
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