# CLASS-10REMINDER AND FACTOR THEOREM - FACTORS OF POLYNOMIAL

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.)