Polynomials: 9 Math
NCERT Exercise2.4: 9th math
Factorization of Polynomials
Factor Theorem
Factor Theorem says that if p(x) is a polynomials of degree `n>=1` and a is any real number, then (i) x – a is a factor of p(x), if p(a) = 0, and (ii) p(a) = 0, if x – a is a factor of p(x).
Proof: By the Remainder Theorem, p(x) = (x – a) q(x) + p(a)
(i) If p(a) = 0, then p(x) = (x – a) q(x), which shows that x – a is a factor of p(x).
(ii) Since x – a is a factor of p(x), p(x) = (x – a) g (x) for same polynomial g(x). In this case, p(x) = (a – a) g(a) = 0
NCERT Exercise 2.4 Polynomials class ninth math Question (1) Determine which of the following polynomials has (x +1) a factor:
(i) x3 + x2 + x + 1
Answer
Let, x + 1 = 0
Therefore, x = –1
Thus, zero of polynomial (x + 1) is equal to –1
Now, let p(x) = x3 + x2 + x + 1
Now, p(–1) = (–1)3 + (–1)2 + (–1) + 1
= –1 + 1 –1 + 1
= 0
Since, p(–1) = 0
Thus, according to Factor Theorem (x + 1) is a factor of given polynomial.
Thus, Answer = yes
(ii) x4 + x3 + x2 + x + 1
Answer
Let, p(x) = x4 + x3 + x2 + x + 1
Then, is x + 1 a factor of p(x) or not?
Here, zeros of x + 1 = –1
Now,
P(–1) = (–1)4 + (–1)3 + (–1)2 + (–1) + 1
= 1 + (–1) + 1 – 1 + 1
= 1 – 1 + 1 – 1 + 1
= 1
Since, here p(x) ≠ 0
Thus, According to Factor Theorem (x + 1) is not a factor of given polynomial.
Thus, Answer = No
(iii) x4 + 3x3 + 3x2 + x + 1
Answer
Let, p(x) = x4 + 3x3 + 3x2 + x + 1
Then, is x + 1 a factor of p(x) or not?
Here, zeros of x + 1 = –1
Now,
p(–1) = (–1)4 + 3(–1)3 + 3(–1)2 + (–1) + 1
= 1 + 3(–1) + 3 × 1 –1 +1
= 1 – 3 + 3
= 1
Since, here p(x) ≠ 0
Thus, According to Factor Theorem (x + 1) is not a factor of given polynomial.
Thus, Answer = No
(iv) x3 – x2 `-(2+sqrt2)x+sqrt2`
Answer
Let, p(x) = x3 – x2 `-(2+sqrt2)x+sqrt2`
Then, is x + 1 a factor of p(x) or not?
Here, zeros of x + 1 = –1
Now, p(–1) = (–1)3 – (–1)2 `-(2+sqrt2)xx(-1)+sqrt2`
= –1 – 1 + 2 +`sqrt2+sqrt2`
= –2 + 2 + `2sqrt2`
`=2sqrt2`
Since, here p(x) ≠ 0
Thus, According to Factor Theorem (x + 1) is not a factor of given polynomial.
Thus, Answer = No
NCERT Exercise 2.4 Polynomials class ninth math Question (2) use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:
(i) p(x) = 2x3 + x2 – 2x – 1, g(x) = x +1
Answer
Given, p(x) = 2x3 + x2 – 2x – 1
And, g(x) = x + 1
Then, find that g(x) is a factor of p(x) or not.
Thus, zero of x + 1 = –1
Now, p(–1) = 2(–1)3 + (–1)2 – 2(–1) – 1
= 2(–1) + 1 + 2 – 1
= –2 + 3 –1
= 0
Since, p(–1) = 0
Thus, According to Factor Theorem g(x) is a factor of given polynomial p(x).
Thus, Answer = Yes
(ii) p(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2
Answer
Given, p(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2
Then, find that g(x) is a factor of p(x) or not.
Now, zero of x + 2 = –2
Now, p(–2) = (–2)3 + 3(–2)2 + 3(–2) + 1
= –8 + 3×4 – 6 + 1
= – 8 + 12 –5
= –1
Here, since p(–2) ≠ 0
Thus, according to Factor Theorem g(x) is not a factor of given polynomial p(x).
Thus, Answer = No
(iii) p(x) = x3 – 4x2 + x + 6, g(x) = x – 3
Answer
Given, p(x) = x3 – 4x2 + x + 6, g(x) = x – 3
Then, find that g(x) is a factor of p(x) or not.
Let, x – 3 = 0
Thus, x = 3
Thus, 3 is the zero of g(x) x – 3
Now, p(3) = x3 – 4x2 + x + 6
= 33 – 4(3)2 + 3 + 6
= 27 – 4 × 9 + 9
= 27 – 36 + 9
= 36 – 36 = 0
Here, since p(3) = 0
Thus, according to Factor Theorem g(x) is a factor of given polynomial p(x).
Thus, Answer = Yes
NCERT Exercise 2.4 Polynomials class ninth math Question (3) Find the value of k, if x – 1 is a factor of p(x) in each of the following cases:
(i) p(x) = x2 + x + k
Answer
Given, p(x) = x2 + x + k
And, x – 1 is a factor of p(x), the k = ?
Since, x – 1 is a factor of p(x) and zeros of x –1 = 1
Thus, p(1) = 0
Now, p(1) = x2 + x + k
⇒ 0 = 12 + 1 + k
⇒ 0 = 1 + 1 + k
⇒ 2 + k = 0
⇒ k = – 2 Answer
(ii) p(x) = 2x2 + kx + `sqrt2`
Answer
Given, p(x) = 2x2 + kx + `sqrt2`
And, x – 1 is a factor of p(x), the k = ?
Since, x – 1 is a factor of p(x) and zeros of x –1 = 1
Thus, p(1) = 0
Now, since, p(x) = 2x2 + kx + `sqrt2`
Thus, p(1) = 2(1)2 + k(1) + `sqrt2`
⇒ 0 = 2 + k + `sqrt2`
⇒ k + `sqrt2` = –2
`=>k=-2-sqrt2` or `-(2+sqrt2)` Answer
(iii) p(x) = kx2 – `sqrt2\x` + 1
Answer
Given, p(x) = kx2 – `sqrt2\x` + 1
And, x – 1 is a factor of p(x), the k = ?
Since, x – 1 is a factor of p(x) and zeros of x –1 = 1
Thus, p(1) = 0
Now, since p(x) = kx2 – `sqrt2\x` + 1
Thus, p(1) = k (1)2 – `sqrt2\xx1` + 1
⇒ 0 = k – `sqrt2` + 1
⇒ k = `sqrt2-1` Answer
(iv) p(x) = kx2 – 3x + k
Answer
Given, p(x) = kx2 – 3x + k
And, x – 1 is a factor of p(x), the k = ?
Since, x – 1 is a factor of p(x) and zeros of x –1 = 1
Thus, p(1) = 0
And, since p(x) = kx2 – 3x + k
Thus, p(1) = k (1)2 – 3 × 1 + k
⇒ 0 = k – 3 + k
⇒ 0 = 2k – 3
⇒ 2k = 3
⇒ k = 3/2 Answer
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