Polynomials: 9 Math
NCERT Exercise 2.4 part-2: 9th math
NCERT Exercise 2.4 Polynomials class ninth math Question (4) Factorize
(i) 12 2 – 7x + 1
Answer
Given, 12 2 – 7x + 1
= 12 x2 – 3x – 4x + 1
= 3x (4x – 1) –1 (4x – 1)
= (4 x – 1) (3x – 1) Answer
(ii) 2 x2 + 7 x + 3
Answer
Given, 2 x2 + 7 x + 3
= 2 x2 + 6 x + x + 3
= 2x (x + 3) + 1 (x + 3)
= (x + 3) (2x + 1) Answer
(iii) 6 x2 + 5x – 6
Answer
Given, 6 x2 + 5x – 6
= 6 x2 + 9 x – 4 x – 6
= 3 x (2x + 3) – 2(2x + 3)
= (2x + 3) (3x – 2) Answer
(iv) 3x2 – x – 4
Answer
Given, 3x2 – x – 4
= 3x2 + 3x – 4x – 4
= 3x (x + 1) – 4( x + 1)
=(x + 1)(3x – 4) Answer
NCERT Exercise 2.4 Polynomials class ninth math Question (5) Factorize
(i) x3 – 2 x2 – x + 2
Answer
Given, x3 – 2 x2 – x + 2
Here constant term = 2
Thus, ab = 2
Now, factor of 2 = 1 and 2
Now, let p(x) = x3 – 2 x2 – x + 2
Thus, p(1) = 13 – 2 (1)2 – 1 + 2
= 1 – 2 – 1 + 2 = 0
i.e. p(1) = 0
Thus, according to Factor Theorem one of the factor of the given polynomial is (x – 1)
Now, x3 – 2 x2 – x + 2
= x3 – x2 –x2 + x – 2x + 2
= x2(x – 1) – x(x –1) – 2(x – 1)
= (x – 1) (x2 – x – 2)
= (x – 1) (x2 – 2x + x – 2)
= (x – 1) [x (x –2) + 1 (x –2)]
= (x – 1) (x – 2) (x + 1)
= (x – 1) (x + 1) (x – 2) Answer
(ii) x3 – 3 x2 – 9 x – 5
Answer
Given, x3 – 3 x2 – 9 x – 5
Here, constant term = –5
i.e. ab = –5
Now, factor of –5 = ±1, ±5
Now, let p(x) = x3 – 3 x2 – 9 x – 5
Thus, p(1) = (1)3 – 3(1)2 – 9 – 5
= 1 – 3 – 9 – 5
= –2 – 9 – 5
= –16
Since, p(1) ≠ 0, thus, this not the zeros of the given polynomial.
And, p(5) = 53 – 3 (5)2 – 9 × 5 – 5
= 125 – 3 × 25 – 45 – 5
= 125 – 75 – 50
= 125 – 125 = 0
Since, p(5) = 0, thus, according to Factor Theorem x – 5 is one of the factor of given polynomial.
Now,
P(x) = x3 – 3 x2 – 9 x – 5
= x3 – 5x2 + 2x2 – 10x + x – 5
= x2(x – 5) + 2x ( x – 5) + 1 (x – 5)
= (x – 5)(x2 + 2x + 1)
= (x – 5) (x2 + x + x + 1)
= (x – 5) [x (x + 1) + 1(x + 1)]
= (x – 5) (x + 1) (x + 1) Answer
(iii) x3 + 13x2 + 32x + 20
Answer
Given, x3 + 13x2 + 32x + 20
Here, constant term ab = 20
And, the factor of constant term 20 = ±1, ±2, ±4, ±5, ±10, ±20
Now, Let, p(x) = x3 + 13x2 + 32x + 20
Now, p(–10) = (–10)3 + 13(–10)2 + 32(–10) + 20
= – 1000 + 13 × 100 – 320 + 20
= – 1000 + 1300 – 300
= – 1300 + 1300 = 0
Now, since p(–10) = 0
Thus, according to Factor Theorem x + 10 is one of the factor of given polynomial
Now, x3 + 13x2 + 32x + 20
= x3 + 10x2 + 3x2 + 30x + 2x + 20
= x2 (x + 10) + 3x( x + 10) + 2( x + 10)
= (x +10) (x2 + 3x + 2)
= (x +10) (x2 + 2x + x + 2)
= (x +10) [x (x + 2) + 1 (x + 2)]
= (x + 10) (x + 2) (x + 1) Answer
(iv) 2y3 + y2 – 2y – 1
Answer
Given, 2y3 + y2 – 2y – 1
Here, constant term ab = 2 × (–1) = –2
Now, factors of –2 = ±1, ±2
Now, let p(y) = 2y3 + y2 – 2y – 1
Thus, p(–1) = 2y3 + y2 – 2y – 1
= 2 (–1)3 + (–1)2 – 2 (–1) – 1
= 2 × (–1) + 1 + 2 – 1
= –2 + 1 + 2 – 1
⇒ p(–1) = 0
Since, p(–1) = 0, thus according to Factor Theorem (y + 1) is one of the factor of given polynomial p(y)
Now, 2y3 + y2 – 2y – 1
= 2 y3 + 2y2 – y2 – y – y – 1
= 2 y2 (y + 1) – y (y + 1) – 1 (y + 1)
= (y +1) (2y2 – y – 1)
= (y + 1) (2 y2 – 2 y + y – 1)
= (y + 1) [2y (y – 1) + 1 (y –1)]
= (y +1) (y –1) (2y + 1) Answer
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