**Factor Theorem**

The **factor
theorem** states that - if f(x) is a polynomial and a is a real number, then
(x – a) is a factor of f(x) if f(a) = 0.

__Proof__**:** If we divide a polynomial f(x) by a binomial x – a,
then we get Q(x) as quotient and R as remainder. Then,

f(x)
= (x – a).Q(x) + R ……………….. (i)

Putting
x = a, then

f(a)
= (a – a).Q(x) + R

or,
f(a) = R

When,
f(a) = 0, then R = 0

Putting
the value of R in (i) we get,

f(x)
= (x – a).Q(x)

It shows that (x – a) is a factor of f(x).

Hence,
(x – a) is a factor of f(x) if f(a) = 0. Proved.

*Corollary 1: If f(x) is a polynomial and a is a real number,
then (x + a) is a factor of f(x) if f(-a) = 0.*

__Proof__**:** If we divide a polynomial f(x) by a binomial x + a,
then we get Q(x) as quotient and R as remainder. Then,

f(x)
= (x + a).Q(x) + R ……………….. (i)

Putting
x = -a, then

f(-a)
= (-a + a).Q(x) + R

or,
f(-a) = R

When,
f(-a) = 0, then R = 0

Putting
the value of R in (i) we get,

f(x)
= (x + a).Q(x)

It shows that (x + a) is a factor of f(x).

Hence,
(x + a) is a factor of f(x) if f(-a) = 0. Proved.

*Corollary 2: If f(x) is a polynomial and a, b are real number,
then (ax - b) is a factor of f(x) if f(b/a) = 0.*

__Proof__**:** If we divide f(x) by ax - b,
then we get Q(x) as quotient and R as remainder. Then,

f(x)
= (ax - b).Q(x) + R ……………… (i)

or, f(x) = a(x – b/a).Q(x) + R

Putting x = b/a, then

f(b/a)
= a(b/a – b/a).Q(x) + R

or,
f(b/a) = R

When,
f(b/a) = 0, then R = 0

Putting
the value of R in (i) we get,

f(x)
= (ax - b).Q(x)

So,
(ax - b) is a factor of f(x).

Hence,
(ax - b) is a factor of f(x) if f(b/a) = 0. Proved.

*Corollary 3: If f(x) is a polynomial and a, b are real number,
then (ax + b) is a factor of f(x) if f(-b/a) = 0.*

__Proof__**:** If we divide f(x) by ax + b,
then we get Q(x) as quotient and R as remainder. Then,

f(x)
= (ax + b).Q(x) + R ……………… (i)

or, f(x) = a(x + b/a).Q(x) + R

Putting
x = -b/a, then

f(-b/a)
= a(-b/a + b/a).Q(x) + R

or,
f(-b/a) = R

When,
f(-b/a) = 0, then R = 0

Putting
the value of R in (i) we get,

f(x)
= (ax + b).Q(x)

So,
(ax + b) is a factor of f(x).

Hence,
(ax + b) is a factor of f(x) if f(-b/a) = 0. Proved.

*Workout
Examples*

*Workout Examples*

*Example 1: Show that (x – 5) is a factor
of 2x ^{2} – 11x + 5.*

*Solution:** Here,*

* f(x) = 2x ^{2} –
11x + 5*

* Comparing x – 5 with x –
a, a = 5*

* **∴** Remainder R = f(5)*

* = 2×5 ^{2} - 11×5 + 5*

* = 50 – 55 + 5*

* = 0*

* **∴** By factor theorem (x – 5) is a factor.*

*Example 2: Show that (x + 3) is a factor
of 2x ^{2} – x - 21.*

*Solution:** Here,*

* f(x) = 2x ^{2} –
x – 21*

* Comparing x + 3 or x - (-3) with x –
a, a = -3*

* **∴** Remainder R = f(-3)*

* = 2×(-3) ^{2} – (-3) – 21 *

* = 18 + 3 – 21 *

* = 0*

* **∴** By factor theorem (x + 3) is a factor.*

*Example 3: Show that (2x – 1) is a factor
of 2x ^{2} – 11x + 5.*

*Solution:** Here,*

* f(x) = 2x ^{2} –
11x + 5*

* Divisor 2x - 1 can be
written as 2(x - 1/2) and comparing with x – a, a = 1/2*

* **∴** Remainder R = f(5)*

* = 2×(1/2) ^{2} - 11×1/2 + 5*

* = 1/2 – 11/2 + 5*

* = 0*

* **∴** By factor theorem (2x – 1) is a factor.*

*Example 4: Show that (2x + 7) is a factor
of 2x ^{3} + 7x^{2} – 4x - 14.*

*Solution:** Here,*

* f(x) = 2x ^{3} +
7x^{2} – 4x – 14 *

* Divisor 2x + 7 can be
written as 2(x + 7/2) and comparing with x – a, a = -7/2*

* **∴** Remainder R = f(-7/2)*

* = 2×(-7/2) ^{3} + 7×(-7/2)^{2}
- 4×(-7/2) – 14 *

* = -343/4 + 343/4 + 14 – 14 *

* = 0*

* **∴** By factor theorem (2x + 7) is a factor.*

*Example 5: If (x – 2) is a factor of the
polynomial x ^{2} – 3x + 5k, find the value of k.*

*Solution:** Here,*

* f(x) = x ^{2} –
3x + 5k*

* Since (x – 2) is a
factor of f(x) = x ^{2} – 3x + 5k,*

* f(2) = 0*

* i.e. 2 ^{2} - 3×2 + 5k = 0*

* or, 4 – 6 + 5k = 0*

* or, -2 + 5k = 0*

* or, 5k = 2*

* or, k
= 2/5*

*Example 6: If (x + 1) and (x – 2) are
factors of x ^{3} + ax^{2} – bx – 6, find the values of a and b.*

*Solution:** Here,*

* f(x) = x ^{3}+ax^{2}–bx–6*

* Since (x + 1) is a
factor of f(x) = x ^{3} + ax^{2} – bx – 6,*

* f(-1) = 0*

* i.e. (-1) ^{3} + a×(-1)^{2} -
b×(-1) – 6 = 0*

* or, -1 + a + b – 6 = 0*

* or, a + b = 7 …………….. (i)*

* Again, since (x – 2) is a factor of **f(x) = x ^{3} + ax^{2} – bx – 6,*

* f(2) = 0*

* i.e. (2) ^{3} + a×(2)^{2} -
b×(2) – 6 = 0*

* or, 8 + 4a - 2b – 6 = 0*

* or, 4a - 2b = -2*

* or, 2a - b
= -1 …………….. (ii)*

* Adding (i)
and (ii), we get,*

* a
+ b + 2a – b = 7 – 1 *

* or, 3a = 6*

* or, a = 6/3 = 2*

* Putting a =
2 in (i), we get,*

* 2
+ b = 7*

* Or, b = 7 – 2 = 5*

* Hence, a = 2
and b = 5.*

*You can comment your
questions or problems regarding the factor
theorem here.*

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