Unit 1.2.3 (B) : Factor theorem
1. (a) State factor theorem.
Solution:
If is a factor of then
OR
If then is a factor of .
1. (b) If is a factor of , what is the degree of quotient.
Solution:
The degree of quotient is .
2. In each of the following, use factor theorem to find whether is a factor of polynomial or not?
(a)
Solution:
Here,
And
Zero of is
is a factor of polynomial
(b)
Solution:
Here,
And
Zero of .
is not a factor of polynomial .
(c)
Solution:
Here,
And
Zero of is 2.
is not a factor of polynomial .
(d)
Solution:
Here,
And
Zero of is .
is a factor of polynomial .
(e)
Solution:
Here,
And
Zero of is .
is not a factor of polynomial .
3. (a) Find the value of , if is a factor of
Solution:
Let,
Zero of is -3.
As is a factor of f(x),
3. (b) Find the value of , if is a factor of
Solution:
Let,
Zero of is -1.
As is a factor of f(x),
3. (c) Find the value of , for which is exactly divisible by
Solution:
Let,
Zero of is .
As is a factor of f(x),
4. Factorize the following by using factor theorem.
(a)
Solution:
Let,
Possible factors of 2 are
is a factor of
Now, using synthetic division, we get,
4(b)
Solution:
Let,
Possible factors of 2 are
is a factor of
Now, using synthetic division, we get,
4(c)
Solution:
Let,
Possible factors of 2 are
is a factor of
Now, using synthetic division, we get,
4(d)
Solution:
Let,
Possible factors of 2 are
is a factor of
Now, using synthetic division, we get,
4(e)
Solution:
Let,
Possible factors of 2 are
is a factor of
Now, using synthetic division, we get,
4(f)
Solution:
Let,
Possible factors of 2 are
is a factor of
Now, using synthetic division, we get,
4(g)
Solution:
Possible factors of 36 are
is a factor of
Now, using synthetic division, we get,
5. Use factor theorem and solve for .
(a)
Solution:
Let
Possible factors of 10 are
is a factor of
Now, using synthetic division, we get,
5(b)
Solution:
Let
Possible factors of 6 are
is a factor of
Now, using synthetic division, we get,
5(c)
Solution:
Let
Possible factors are
is a factor of
Now, using synthetic division, we get,
5(d)
Solution:
Let,
Possible factors of 5 are
is a factor of
Now using synthetic division, we get,
5(e)
Solution:
Let,
Possible factors of 24 are
is a factor of
Now, using synthetic division, we get,
5(f)
Solution:
Given,
Possible factors of 6 are
is a factor of
Now, using synthetic division, we get,
4(b)
Solution:
Let,
Possible factors of 2 are
is a factor of
Now, using synthetic division, we get,
4(c)
Solution:
Let,
Possible factors of 2 are
is a factor of
Now, using synthetic division, we get,
4(d)
Solution:
Let,
Possible factors of 2 are
is a factor of
Now, using synthetic division, we get,
4(e)
Solution:
Let,
Possible factors of 2 are
is a factor of
Now, using synthetic division, we get,
4(f)
Solution:
Let,
Possible factors of 2 are
is a factor of
Now, using synthetic division, we get,
4(g)
Solution:
Possible factors of 36 are
is a factor of
Now, using synthetic division, we get,
5. Use factor theorem and solve for .
(a)
Solution:
Let
Possible factors of 10 are
is a factor of
Now, using synthetic division, we get,
5(b)
Solution:
Let
Possible factors of 6 are
is a factor of
Now, using synthetic division, we get,
5(c)
Solution:
Let
Possible factors are
is a factor of
Now, using synthetic division, we get,
5(d)
Solution:
Let,
Possible factors of 5 are
is a factor of
Now using synthetic division, we get,
5(e)
Solution:
Let,
Possible factors of 24 are
is a factor of
Now, using synthetic division, we get,
5(f)
Solution:
Given,
Possible factors of 6 are
is a factor of
Now, using synthetic division, we get,
1. (a) State factor theorem.
Solution:
If is a factor of then
OR
If then is a factor of .
1. (b) If is a factor of , what is the degree of quotient.
Solution:
The degree of quotient is .
2. In each of the following, use factor theorem to find whether is a factor of polynomial or not?
(a)
Solution:
Here,
And
Zero of is
is a factor of polynomial
(b)
Solution:
Here,
And
Zero of .
is not a factor of polynomial .
(c)
Solution:
Here,
And
Zero of is 2.
is not a factor of polynomial .
(d)
Solution:
Here,
And
Zero of is .
is a factor of polynomial .
(e)
Solution:
Here,
And
Zero of is .
is not a factor of polynomial .
3. (a) Find the value of , if is a factor of
Solution:
Let,
Zero of is -3.
As is a factor of f(x),
3. (b) Find the value of , if is a factor of
Solution:
Let,
Zero of is -1.
As is a factor of f(x),
3. (c) Find the value of , for which is exactly divisible by
Solution:
Let,
Zero of is .
As is a factor of f(x),
4. Factorize the following by using factor theorem.
(a)
Solution:
Let,
Possible factors of 2 are
is a factor of
Now, using synthetic division, we get,
4(b)
Solution:
Let,
Possible factors of 2 are
is a factor of
Now, using synthetic division, we get,
4(c)
Solution:
Let,
Possible factors of 2 are
is a factor of
Now, using synthetic division, we get,
4(d)
Solution:
Let,
Possible factors of 2 are
is a factor of
Now, using synthetic division, we get,
4(e)
Solution:
Let,
Possible factors of 2 are
is a factor of
Now, using synthetic division, we get,
4(f)
Solution:
Let,
Possible factors of 2 are
is a factor of
Now, using synthetic division, we get,
4(g)
Solution:
Possible factors of 36 are
is a factor of
Now, using synthetic division, we get,
5. Use factor theorem and solve for .
(a)
Solution:
Let
Possible factors of 10 are
is a factor of
Now, using synthetic division, we get,
5(b)
Solution:
Let
Possible factors of 6 are
is a factor of
Now, using synthetic division, we get,
5(c)
Solution:
Let
Possible factors are
is a factor of
Now, using synthetic division, we get,
5(d)
Solution:
Let,
Possible factors of 5 are
is a factor of
Now using synthetic division, we get,
5(e)
Solution:
Let,
Possible factors of 24 are
is a factor of
Now, using synthetic division, we get,
5(f)
Solution:
Given,
Possible factors of 6 are
is a factor of
Now, using synthetic division, we get,
4(b)
Solution:
Let,
Possible factors of 2 are
is a factor of
Now, using synthetic division, we get,
4(c)
Solution:
Let,
Possible factors of 2 are
is a factor of
Now, using synthetic division, we get,
4(d)
Solution:
Let,
Possible factors of 2 are
is a factor of
Now, using synthetic division, we get,
4(e)
Solution:
Let,
Possible factors of 2 are
is a factor of
Now, using synthetic division, we get,
4(f)
Solution:
Let,
Possible factors of 2 are
is a factor of
Now, using synthetic division, we get,
4(g)
Solution:
Possible factors of 36 are
is a factor of
Now, using synthetic division, we get,
5. Use factor theorem and solve for .
(a)
Solution:
Let
Possible factors of 10 are
is a factor of
Now, using synthetic division, we get,
5(b)
Solution:
Let
Possible factors of 6 are
is a factor of
Now, using synthetic division, we get,
5(c)
Solution:
Let
Possible factors are
is a factor of
Now, using synthetic division, we get,
5(d)
Solution:
Let,
Possible factors of 5 are
is a factor of
Now using synthetic division, we get,
5(e)
Solution:
Let,
Possible factors of 24 are
is a factor of
Now, using synthetic division, we get,
5(f)
Solution:
Given,
Possible factors of 6 are
is a factor of
Now, using synthetic division, we get,