Factor Theorem

Factor Theorem

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

Example 1: Show that (x – 5) is a factor of 2x2 – 11x + 5.

Solution: Here,

                        f(x) = 2x2 – 11x + 5

                        Comparing x – 5 with x – a, a = 5

                        Remainder R = f(5)

                                               = 2×52 - 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 2x2 – x - 21.

Solution: Here,

                        f(x) = 2x2 – 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 2x2 – 11x + 5.

Solution: Here,

                        f(x) = 2x2 – 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 2x3 + 7x2 – 4x - 14.

Solution: Here,

                        f(x) = 2x3 + 7x2 – 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 x2 – 3x + 5k, find the value of k.

Solution: Here,

                        f(x) = x2 – 3x + 5k

                        Since (x – 2) is a factor of f(x) = x2 – 3x + 5k,

                         f(2) = 0

            i.e.       22 - 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 x3 + ax2 – bx – 6, find the values of a and b.

Solution: Here,

                        f(x) = x3+ax2–bx–6

                        Since (x + 1) is a factor of f(x) = x3 + ax2 – 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) = x3 + ax2 – 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.

 

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