Chapter 2. Polynomials
Exercise 2.4
Q.1. Determine which of the following polynomials has (x+ 1) a factor:
(i) x3 +x2 + x +1 (ii) x4 + x3 + x2 + x +1
(iii) x4 + 3x3 + 3x2 + x +1
Solution:
(i) If (x + 1) is a factor of p(x) = x3+ x2+ x + 1, then p (−1) must be zero, otherwise (x + 1) is not a factor of p(x).
P(x) = x3 + x2 + x + 1
P(−1)= (−1)3 + (−1)2 + (−1) + 1
= − 1 + 1 − 1 − 1
= 0
∵ P(x) = 0
Hence, x + 1 is a factor of this polynomial.
(ii) If (x + 1) is a factor of p(x) = x4+ x3+ x2+ x + 1, then p (−1) must be zero, Otherwise (x + 1) is not a factor of p(x).
P(x) = x4+ x3+ x2+ x + 1
P(−1) = (−1)4 + (−1)3 + (−1)2 + (−1) + 1
= 1 − 1 + 1 −1 + 1
= 1
As P(x) ≠ 0, (− 1)
Therefore, x + 1 is not a factor of this polynomial.
(iii) If (x + 1) is a factor of polynomial p(x) = x4+ 3x3+ 3x2+ x + 1, then p (−1) must be 0, otherwise (x + 1) is not a factor of this polynomial.
P(x) = x4+ 3x3+ 3x2+ x + 1
P(−1) = (−1)4+ 3(−1)3+ 3(−1)2+ (−1) + 1
= 1 − 3 + 3 − 1 + 1
= 1
As P(x) ≠ 0, (−1)
Therefore (x+1) is not a factor of this polynomial .
Q.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
(ii) P(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2
(iii) P(x) = x3 − 4 x2 + x + 6, g(x) = x − 3
Solution:
(i) If g(x) = x + 1 is a factor of the given polynomial p(x), then p (−1) must be zero.
P (x) = 2x3 + x2 − 2x − 1
P (−1) = 2(−1)3+ (−1)2− 2(−1) − 1
= 2(−1) + 1 + 2 – 1
= 0
∵ P(x) = 0
Hence, g(x) = x + 1 is a factor of the given polynomial.
(ii) If g(x) = x + 2 is a factor of the given polynomial p(x), then p (−2) must be 0.
P (x) = x3+3x2+ 3x + 1
P (−2) = (−2)3+ 3(−2)2+ 3(−2) + 1
= − 8 + 12 − 6 + 1
= −1
As P(x) ≠ 0,
Hence, g(x) = x + 2 is not a factor of the given polynomial.
(iii) If g(x) = x − 3 is a factor of the given polynomial p(x), then P(3) must be 0.
P(x) = x3− 4 x2+ x + 6
P(3) = (3)3 − 4(3)2 + 3 + 6
= 27 −36 + 9
= 0
Hence, g(x) = x − 3 is a factor of the given polynomial.
Solution:
: If x − 1 is a factor of polynomial p(x), then P(1) must be 0.
(i) P(x) = x2+ x + k
P(1) = (1)2+ 1 + k
= 1+1+
= 2+k
k =−2
(iv) P(x) = kx2-3x + k
P(1) = k(1)2-3(1) + k
= k-3+k
2k = 3
K = 3/2
Question 4: Factorise:
(i) 12x2− 7x + 1 (ii) 2x2+ 7x + 3
(iii) 6x2+ 5x – 6 (iv) 3x2− x − 4
Solution:
(i) 12x2− 7x + 1 we can find two numbers,
Such that pq = 12 × 1 = 12 and p + q = −7.
They are p = −4 and q = −3
Here, 12x2− 7x + 1
= 12x2− 4x − 3x + 1
= 4x (3x − 1) − 1 (3x − 1)
= (3x − 1) (4x − 1)
(ii) 2x2+ 7x + 3 we can find two numbers such that pq = 2 × 3= 6 and p + q = 7.
They are p = 6 and q = 1.
Here, 2x2 + 7x + 3
= 2x2+ 6x + x + 3
= 2x (x + 3) + 1 (x + 3)
= (x + 3) (2x+ 1)
(iii) 6x2+ 5x − 6 we can find two numbers such that pq = −36 and p + q = 5.
They are p = 9 and q = −4.
Here, 6x2+ 5x – 6
= 6x2+ 9x − 4x – 6
= 3x (2x + 3) − 2 (2x + 3)
= (2x + 3) (3x − 2)
(iv) 3x2− x − 4 we can find two numbers,
such that pq = 3 × (−4) = −12 and p + q = −1.
They are p = −4 and q = 3
Here, 3x2− x − 4
= 3x2− 4x + 3x – 4
= x (3x − 4) + 1 (3x − 4)
= (3x − 4) (x + 1)
Question 5. Factorize:
(i) x3− 2x2− x + 2 (ii) x3+ 3x2−9x − 5
(iii) x3+ 13x2+ 32x + 20 (iv) 2y3+ y2− 2y – 1
Solution:
(i) Let P(x) = x3− 2x2− x + 2 all the factor are there. These are ± 1, ± 2.
By trial method, P (1) = (1)3− 2(1)2− 1 + 2
= 1 − 2 − 1+ 2
= 0 Therefore, (x − 1) is factor of polynomial p(x)
Let us find the quotient on dividing x3− 2x2− x + 2 by x − 1.
By long division method
Now,
Dividend = Divisor × Quotient + remainder
x3− 2x2− x + 2 = (x – 1) ( X2– x – 2) + 0
= (x – 1) (x2–2x+x–2)
= (x – 1) [x (x–2) + 1(x–2)]
= (x – 1) (x + 1) (x – 2)
(ii) Let p(x) = x3 – 3x2−9x – 5 all the factor are there. These are ± 1, ± 2.
By trial method, p (–1) = (–1)3– 3(1)2− 9(1) – 5
= –1– 3–9–5 =0
Therefore (x+1) is the factor of polynomial p(x).
. Let us find the quotient on dividing x3– 3x2−9x – 5 by x+1.
By long division method
Now,
Dividend = Divisor × Quotient + remainder
x3– 3x2−9x – 5 = (x +1) ( X2–4x – 5) + 0
=(x + 1) (x2–5x+x–5)
=(x + 1) [x (x–5) +1(x–5)]
=(x + 1) (x + 1) (x – 5)
(iii) Let p(x) = x3+ 13x2+ 32x + 20 all the factor are there.
These are ± 1, ± 2, ± 3, ± 4.
By trial method, p (–1) = (–1)3+13(–1)2+ 32(–1) +20
= –1+13–32+20 = 0
Therefore (x+1) is the factor of polynomial p(x).
Let us find the quotient on dividing x3+ 13x2+ 32x + 20 by x+1
By long division method
Now,
Dividend = Divisor × Quotient + remainder
x3 +13x2 + 32x + 20 = (x +1) ( x2 + 12x + 20) + 0
=(x + 1) (x2+10x+2x+20)
=(x + 1) [x (x+10) +2(x+10)]
=(x + 1) (x + 2) (x + 10)
(iv) Let p(y) = 2y3+ y2− 2y – 1 all the factor are there. These are ± 1, ± 2.
By trial method, p (1) =2(1)3 + (1)2 – 2(1) – 1
=2 + 1 – 2 – 1 =0
Therefore (y–1) is the factor of polynomial p(y).
Let us find the quotient on dividing 2y3+ y2− 2y – 1 by y–1.
By long division method
Now,
Dividend = Divisor × Quotient + remainder
2y3+ y2− 2y −1 =(y − 1) (2y2+3y + 1)
= (y − 1) (2y2+2y
= (y − 1) [2y (y+1) + 1 (y + 1)]
= (y − 1) (y + 1) (2y + 1)