Factor Theorem for Polynomials :
In this section, you will learn how factor theorem can be used for polynomials.
Factor Theorem :
If p(x) is a polynomial of degree n ≥ 1 and ‘a’ is any real number then
(i) p(a) = 0 implies (x - a) is a factor of p(x).
(ii) (x - a) is a factor of p(x) implies p(a) = 0.
Note :
(i) (x-a) is a factor of p(x), if p(a) = 0.
(ii) (x+a) is a factor of p(x), if p(-a) = 0.
(iii) (ax+b) is a factor of p(x), if p(-b/a) = 0.
(iv) (x-a)(x-b) is a factor of p(x), if p(a) = 0 and p(b) = 0.
Question 1 :
Determine the value of m, if (x + 3) is a factor of
x3 - 3x2 - mx + 24
Solution :
Let
f(x) = x3 - 3x2 - mx + 24
Equate the factor (x + 3) to zero.
x + 3 = 0
Solve for x.
x = -3
By factor theorem,
(x + 3) is factor of f(x), if f(-3) = 0
Then,
f(-3) = 0
(-3)3 - 3(-3)2 - m(-3) + 24 = 0
-27 - 3(9) + 3m + 24 = 0
-27 - 27 + 3m + 24 = 0
3m - 30 = 0
3m = 30
m = 10
Question 2 :
If both (x - 2) and (x - 1/2) are the factors of
ax2 + 5x + b,
then show that a = b.
Solution :
Let p(x) = ax2 + 5x + b
Equate the factor (x - 2) to zero.
x - 2 = 0
Solve for x.
x = 2
By factor theorem,
(x - 2) is factor of p(x), if p(2) = 0
Then,
p(2) = 0
a(2)2 + 5(2) + b = 0
4a + 10 + b = 0
b = -4a - 10 -----(1)
Equate the factor (x - 1/2) to zero.
x - 1/2 = 0
Solve for x.
x = 1/2
By factor theorem,
(x - 1/2) is factor of p(x), if p(1/2) = 0
p(1/2) = 0
a(1/2)2 + 5(1/2) + b = 0
a/4 + 5/2 + b = 0
a/4 + 10/4 + 4b/4 = 0
(a + 10 + 4b) / 4 = 0
a + 10 + 4b = 0
From (1), substitute (-4a - 10) for b.
a + 10 + 4(-4a - 10) = 0
a + 10 - 16a - 40 = 0
-15a - 30 = 0
-15a = 30
a = -2 -----(2)
Substitute -2 for a in (1).
(1)-----> b = -4(-2) - 10
b = 8 - 10
b = -2 -----(3)
From (2) and (3), we have
a = b
Question 3 :
If (x - 1) divides the polynomial
kx3 - 2x2 + 25x - 26
without remainder, then find the value of k .
Solution :
Let
p(x) = kx3 - 2x2 + 25x - 26
Because (x - 1) divides p(x) without remainder, (x - 1) is a factor of p(x).
Equate the factor (x - 1) to zero.
x - 1 = 0
Solve for x.
x = 1
By factor theorem,
(x - 1) is factor of p(x), if p(1) = 0
Then,
p(1) = 0
k(1)3 - 2(1)2 + 25(1) - 26 = 0
k - 2 + 25 - 26 = 0
k - 28 + 25 = 0
k - 3 = 0
k = 3
Question 4 :
Check if (x + 2) and (x - 4) are the sides of a rectangle whose area is x2 - 2x - 8 by using factor theorem.
Solution :
If (x + 2) and (x - 4) are the sides of a rectangle, then the area of the rectangle is
(x + 2)(x - 4)
But, it is given that the area of the rectangle is
x2 - 2x - 8
Then,
x2 - 2x - 8 = (x + 2)(x - 4)
From, the above equation, (x + 2) and (x - 4) must be the factors of x2 - 2x - 8.
Let's verify.
Let p(x) = x2 - 2x - 8
By factor theorem,
(x + 2) is factor of p(x), if p(-2) = 0
(x - 4) is factor of p(x), if p(-2) = 0
Then,
p(-2) = (-2)2 - 2(-2) - 8 p(-2) = 4 + 4 - 8 p(-2) = 0 |
p(4) = 42 - 2(4) - 8 p(4) = 16 - 8 - 8 p(4) = 0 |
So, (x + 2) and (x - 4) are the factors of (x2 - 2x - 8).
Hence, (x + 2) and (x - 4) are the sides of the rectangle.
After having gone through the stuff given above, we hope that the students would have understood how factor theorem can be used for polynomials.
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