Factor Theorem Applications| Factor theorem example problems
Statement and proof of Factor Theorem
Factor theorem practice problems
Example – 1 : Find the value of k, if (2x-3) is a factor of 2x3 – 9x2 + x + k
Solution: Let f(x) = 2x3 – 9x2 + x + k & g(x) = 2x-3
Since g(x) is a factor of f(x)
This implies that f(3/2) = 0
Now put x = 3/2 in f(x)
-48/4 + K = 0
K = 12
Example – 2 :Find the value of “n” and ‘m” such that the polynomial x3 – nx2– 13x + m has (x + 1) and ( x – 2) as factor.
Solution: Let f(x) =x3 – nx2– 13x + m
Since (x +1) is a factor of f(x)
i.e f(-1) = 0
So f ( -1) = – 1 -n +13 +m = 0
m -n = -12 – – – – – – – – ( 1)
Since (x -2) is a factor of f(x)
i.e f(2) = 0
So f ( 3) = 8 -4n – 26 + m = 0
m -4n = 18 – – – – – – – – – (2)
Now from equations (1) and (2)
n = -10 & m = -22
Example – 3 : What much be added to x3 – 3x2 + 4x – 15 to obtain a polynomial which is exactly divisible by x-3 ?
Solution: Let f (x) = x3 – 3x2 + 4x – 15 & g(x) = x -3
Since the degree of the g(x) is 1. So the degree of the remainder is 0
Let “b” be added to f(x), so that it may be exactly divisible by g(x)
i.e f(x) = (x3 – 3x2 + 4x -15) + b
Since f(x) is exactly divisible by (x-3)
So f(3) = 0
27 – 27 + 12 – 15 + b = 0
b = 3
i. e g(x) is a factor of f(x) while add constant of 3 to f(x)
Application of the factor theorem
Application 1
Prove that (x- a) is a factor of xn – an for any natural number of “n” by using factor theorem
Let f(x) = xn – an , g(x) = x –a Where n = Natural number
The remainder when f(x) is divide by g(x) is f(a)
Now put x = a in f(x) then
f(a) = an – an = 0
So f(a) = 0 for any natural number of “n”
i.e (x- a) is a factor of xn – an where n N
Application 2
Prove that (x + a) is a factor of xn + an for any odd number of “n” by using factor theorem
Let f(x) = xn + an , g(x) = x + a Where n = odd number
The remainder when f(x) is divide by g(x) is f(-a)
Now put x = – a in f(x) then
f(-a) = (-a)n + an = – an + an = 0
So f(-a) = 0 for any odd number of “n”
i.e (x + a) is a factor of xn + an where n = odd number
Example – 4 : Find the factors of the following polynomials
i) a3 +8 ii) a6 – b6 iii) 125a3 + 216b3
Solutions :
i ) a3 +8 = a3 + 23 where exponential is odd number
According to Application 2
(a +2 ) is factor of a3 + 23
ii ) a6 – b6 where exponential is natural number
According to Application 1
(a – b ) is factor of a6 – b6
iii ) 125a3 + 216b3 = (5a)3 + (6b)3 where exponential is odd number
According to Application 2
(5a + 6b ) is factor of 125a3 + 216b3
Application 3 (Factorisation of a quatratic polynomial)
Factorise any quadratic polynomial ax2 + bx + c ( where a 0 and a,b & c are constants ), we have to write b as the sum of two numbers whose product is “ac”.
Proof:
Here a polynomial ax2 + bx + c ( where a 0 and a,b & c are constants )
Let its factors be (rx + s) & (px + q) then
ax2 + bx + c = (rx + s) (px + q) = rpx2 + (sp + rq)x + sq.
Comparing the coefficients of x2, we get that a = rp
Similarly comparing the coefficient of x, we get b = sp + rq
And on comparing the constant term, we get c = sq.
This shows that “b” is the sum of two numbers sp & rq, whose product is
(sp) (rp) =(rp) (sq) = a c
Therefore, to factorise ax2 + bx + c , we have to write ‘ b ‘ as the sum of two numbers whose product is ‘ ac ‘
Example – 5 : Factorise using factor theorem 6x2 + 17x +5
Solution: By using the application -3
If we can find two numbers p and q such that p +q = 17 and pq = 6 x 5 = 30, we can get the factors
So, let us look for the factors of 30. Some are
1 & 30
3 & 10
2 & 15
Of these pairs 2 & 15 will give us p +q = 2 + 15 = 17
So, 6x2 + 17x +5 = 6x2 + (15 +2)x +5
= 2x(3x +1) + 5(3x + 1)
= (2x +5) (3x +1)
Therefore (2x +5) & (3x +1) are factors of a polynomial 6x2 + 17x +5
Example – 6 : Factorise using factor theorem 2x2 + 5x + 3
Solution: By using the application -3
If we can find two numbers p and q such that p +q = 5 and pq = 2 x 3 = 6, we can get the factors
So, let us look for the factors of 6. Some are
1 & 6
2 & 3
Of these pairs 2 & 3 will give us p +q = 2 +3 = 5
So, 2x2 + 5x + 3 = 2x2 + (2+3)x + 3
= x(2x +3) + (2x + 3)
= (2x +3) (x +1)
Therefore (2x +3) & (x +1) are factors of a polynomial 2x2 + 5x + 3
Solution: Let p(x) = x4 +4x3 -x2 – 16x -12
The constant term of p(x) is -12 and its factors are
±1 , ±2, ±3, ±4, ±6 & ±12
So in beginning – – – – – – till the value of p(x) os zero
Put x = 1 in p(x)
P(1) = 1 + 4 – 1 – 16 – 12 = 24 ≠ 0
i.e (x – 1) is not a factor for p(x)
Now put x = -1 in p(x)
P(-1) = 1 -4 -1 +16 – 12 = 17 – 17 = 0
i.e (x + 1) is a factor for p(x)
By dividing p(x) with (x+1),
We get the quotient x3 + 3 x2 – 4x – 12
So p (x) = (x +1) (x3 + 3 x2 – 4x – 12)
Let q(x) = x3 + 3 x2 – 4x – 12
Again we should starts to find q(2), q(-2) – – – – –
q(2) = 8 + 12 – 8 12 = 0
Therefore ( x- 2) is a factor of q(x)
By dividing q(x) with (x -2), we get quotient x2 + 5x + 6
q(x) = (x -2) (x2 + 5x + 6)
Finally
P(x) = x4 +4x3 -x2 – 16x -12
= (x +1) (x3 + 3 x2 – 4x – 12)
= (x +1) (x -2) (x2 + 5x + 6)
= (x +1) (x -2) (x2 + 3 x +2x + 6)
= (x +1) (x -2) [ x(x + 3) + 2(x + 3)]
= (x +1) (x +2) (x -2) (x + 3)
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