Ncert Solutions Class 9 Math Chapter 2 Polynomials Ex 2 4
Ncert Solutions Class 9 Math Chapter 2 Polynomials Ex 2 4
Ncert Solutions Class 9 Math Chapter 2 Polynomials Ex 2 4
Reasoning:
When a polynomial p(x) is divided by x – a and if p(a) = 0 then (x – a) is a factor of
p(x). The root of x+1=0 is –1.
Solution:
(i) Let p( x) = x + x + x + 1
3 2
(ii) p( x) = x 3 + 3x 2 + 3x + 1, g ( x) = x + 2
(iii) p( x) = x3 − 4 x 2 + x + 6, g ( x) = x − 3
Reasoning:
By factor theorem, (x – a) is a factor of a polynomial p(x) if p(a) = 0.
To find if g(x )= x+a is a factor of p(x), we need to find the root of g(x).
x + a = 0 → x = –a
Solution:
(i) Let p( x) = 2 x + x − 2 x − 1, g ( x) = x + 1
3 2
x + 1 = 0 → x = –1
Now,
p (−1) = 2(−1)3 + (−1) 2 − 2(−1) − 1
= −2 + 1 + 2 − 1
=0
(ii) Let p( x) = x + 3x + 3x + 1, g ( x) = x + 2
3 2
x+2=0 → x = –2
Now,
p (−2) = (−2)3 + 3(−2) 2 + 3(−2) + 1
= −8 + 12 − 6 + 1
= −1 0
Since the remainder of p(−2) 0 , by factor theorem we can say g(x) = x+2 is not a
factor of p( x) = x + 3x + 3x + 1.
3 2
(iii) Let p( x) = x − 4 x + x + 6, g ( x) = x − 3
3 2
x − 3 = 0 → x =3
Now,
p (3) = (3)3 − 4(3) 2 + 3 + 6
= 27 − 36 + 3 + 6
=0
Since the remainder of p(3) = 0 , by factor theorem we can say g(x) = x-3 is a factor of
p( x) = x 3 − 4 x 2 + x + 6.
Q3. Find the value of k, if x – 1 is a factor of p(x) in each of the following cases:
(iii) p( x) = kx 2 − 2 x + 1 (iv) p( x) = kx 2 − 3x + k
Reasoning:
By factor theorem, if x-1 is a factor of p(x), then p (1) = 0 .
Solution:
(i)
p ( x) = x 2 + x + k
p (1) = (1) 2 + (1) + k
0=2+k
k = −2
(ii)
p ( x) = 2 x 2 + kx + 2
p (1) = 2(1) 2 + k (1) + 2
0=2+k + 2
k = −(2 + 2)
(iii)
p( x) = kx 2 − 2 x + 1
p(1) = k (1) 2 − 2(1) + 1
0 = k − 2 +1
k = 2 −1
(iv)
p( x) = kx 2 − 3x − k
p(1) = k (12 ) − 3(1) − k
0 = 2k − 3
3
k =
2
Q4. Factorise:
(i) 12 x 2 − 7 x + 1 (ii) 2 x 2 + 7 x + 3
Reasoning:
By splitting method, we can find factors using the following method.
Find 2 numbers p, q such that:
i. p + q = co-efficient of x
ii. pq = co-efficient of x2 and the constant term.
Solution:
(i) 12 x − 7 x + 1
2
p + q = −7 (co-efficient of x)
pq = 12 1 = 12 (co-efficient of x 2 and the constant term.)
(ii) 2 x + 7 x + 3
2
p + q = 7 (co-efficient of x)
pq = 2 3 = 6 (co-efficient of x 2 and the constant term.)
(iii) 6x2 + 5x − 6
p + q = 5 (co-efficient of x)
pq = 6 ( −6 ) = −36 (co-efficient of x 2 and the constant term.)
6 x2 + 5x − 6 = 6 x2 + 9 x − 4 x − 6
= 3x(2 x + 3) − 2(2 x + 3)
= (3x − 2)(2 x + 3)
(iv) 3x2 − x − 4
p + q = −1 (co-efficient of x)
Q5. Factorise:
(i) x3 − 2 x2 − x + 2 (ii) x3 − 3x2 − 9 x − 5
Solution:
(i) Let p( x) = x − 2 x − x + 2
3 2
Method 2:
x3 − 2 x 2 − x + 2 = ( x 3 − 2 x 2 ) − ( x − 2)
= x 2 ( x − 2) − 1( x − 2)
= ( x − 2)( x 2 − 1)
= ( x − 2)( x + 1)( x − 1)
(By using a 2 − b 2 = (a + b)(a − b) )
(ii) Let p( x) = x − 3x − 9 x − 5
3 2
x3 + x 2
− 4 x2 − 9 x
− 4 x2 − 4 x
− 5x − 5
− 5x − 5
0
Hence x − 3x − 9 x − 5 = ( x + 1)( x − 4 x − 5)
3 2 2
Now taking x − 4 x − 5, find 2 numbers p, q such that:
2
i. p + q = co-efficient of x
ii. pq = co-efficient of x2 and the constant term.
p + q = −4 (co-efficient of x)
pq = 1 −5 = −5 (co-efficient of x 2 and the constant term.)
= ( x + 1) 2 ( x − 5)
We shall find a factor of p(x) by using some trial value of x, say x = -1. (Since all the
terms are positive.)
p(−1) = (−1)3 + 13(−1) 2 + 32(−1) + 20
= −1 + 13 − 32 + 20
=0
Since the remainder of p(−1) = 0 , by factor theorem we can say x=-1 is a factor of
p( x) = x 3 + 13x 2 + 32 x + 20.
x 2 + 12 x + 20
x + 1 x3 + 13x 2 + 32 x + 20
x3 + x 2
12 x 2 + 32 x
12 x 2 + 12 x
20 x + 20
20 x + 20
0
x + 13x + 32 x + 20 = ( x + 1)( x 2 + 12 x + 20)
3 2
Now taking x2 + 12 x + 20, find 2 numbers p, q such that:
i. p + q = co-efficient of2x
ii. pq = co-efficient of
x and the constant term.
p + q = 12 (co-efficient of x) 2
pq = 1 20 = 20 (co-efficient of x and the constant term.)
By trial and error method, we get p = 10, q = 2.
Now splitting the middle term of the given polynomial,
x2 + 12x + 20 = x2 + 10x + 2x + 20
= x( x + 10) + 2( x + 10)
= ( x + 10)( x + 2)
x + 13x + 32x + 20 = ( x + 1)( x + 10)( x + 2)
3 2
Method 2:
x3 + 13x2 + 32x + 20 = x3 + 10x2 + 3x2 + 30x + 2x + 20
= x 2 ( x + 10) + 3x( x + 10) + 2( x + 10)
= ( x + 10)( x 2 + 3x + 2)
= ( x + 10)( x 2 + 2 x + x + 2)
= ( x + 10)[ x( x + 2) + 1( x + 2)]
= ( x + 10)( x + 2)( x + 1)
(iv) Let p( y ) = 2 y + y − 2 y − 1
3 2