2. Write the degree of the following polynomials. (a) 3 +7x2 - 4x + x2 (b)6xy-7x²y+8x²y² (c) 4x - 3x'y + 3xy+ 8y5 (d) x³y³-8x²y²+7xy+8
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Answers
Answer:
Exercise– 2.1
1. Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.
Ans. (i) 4x2 – 3x + 7
⇒ 4x2 – 3x + 7x°
∵ All the exponents of x are whole numbers.
∴ 4x2 – 3x + 7 is a polynomial in one variable.
(ii)
∵ All the exponents of y are whole numbers.
∴ is a polynomial in one variable.
(v) x10 + y3 + t50
∵; Exponent of every variable is a whole number,
∴ x10 + y3 + t50 is a polynomial in x, y and t, i.e. in three variables.
2. Write the co-efficients of x2 in each of the following:
(i) 2 + x2 + x
(ii) 2 – x2 + x3
(iii)
(v)
Ans. (i) 2 + x2 + x
The co-efficient of x2 is 1.
(ii) 2 – x2 + x3
The co-efficient of x2 is (–1).
(iii)
The co-efficient of x2 is
(iv)
∴ The co-efficient of x2 is 0
3. Give one example each of a binomial of degree 35, and of a monomial of degree 100.
Ans. (i) A binomial of degree 35 can be: 3x35 – 4
(ii) A monomial of degree 100 can be:
4. Write the degree of each of the following polynomials:
(i) 5x3 + 4x2 + 7x (ii) 4 - y2 (iii) (iv) 3
Ans. (i) 5x3 + 4x2 + 7x
∵ The highest exponent of x is 3.
∴ The degree of the polynomial is 3.
(ii) 4 – y2
∵ The highest exponent of y is 2.
∴ The degree of the polynomial is 2.
(iii)
∵ The highest exponent of t is 1.
∴ The degree of the polynomial is 1.
(iv) 3
since, 3 = 3x°
∴ The degree of the polynomial 3 is 0.
5. Classify the following as linear, quadratic and cubic polynomials:
(i) x2 + x
(ii) x – x3
(iii) y + y2 + 4
(iv) 1 + x
(v) 3t
(vi) r2
(vii) 7x3
Ans. (i) x2 +x
∵ The degree of x2 + x is 2.
∴ It is a quadratic polynomial.
(ii) x – x3
∵ The degree of x – x3 is 3.
∴ It is a cubic polynomial.
(iii) y + y2 + 4
∵ The degree of y + y2 + 4 is 2.
∴ It is a quadratic polynomial.
(iv) 1 + x
∵ The degree of 1 + x is 1.
∴ It is a linear polynomial.
(v) 3t
∵ The degree of 3t is 1.
∴ It is a linear polynomial.
(vi) r2
∵ The degree of r2 is 2.
∴ It is a quadratic polynomial.
(vii) 7x3
∵ The degree of 7x3 is 3.
∴ It is a cubic polynomial.
Exercise– 2.2
1. Find the value of the polynomial 5x – 4x2 + 3 at
(i) x = 0 (ii) x = –1 (iii) x = 2
Ans. (i) ∵ p(x) = 5x – 4x2 + 3 = 5(x) – 4(x)2 + 3
∴ p(0) = 5(0) – 4(0) + 3 = 0 – 0 + 3 = 3
Thus, the value of 5x – 4x2 + 3 at x = 0 is 3.
(ii) ∵ p(x) = 5x – 4x2 + 3 = 5(x) – 4(x)2 + 3
∴ p(–1) = 5(–1) – 4(–1)2 + 3 = – 5 – 4(1) + 3
= –5 – 4 + 3 = –9 + 3 = –6
∴ The value of 5x – 4x2 + 3 at x = –1 is –6.
(iii) ∵ p(x) = 5x – 4x2 + 3 = 5(x) – 4(x)2 + 3
∴ p(2) = 5(2) – 4(2)2 + 3 = –10 – 4(4) + 3
= 10 – 16 + 3 = –3
Thus the value of 5x – 4x2 + 3 at x = 2 is –3
2. Find p(0), p(1) and p(2) for each of the following polynomials:
(i) p(y) = y2 – y + 1
(ii) p(t) = 2 + t + 2t2 – t3
(iii) p(x) = x3
(iv) p(x) = (x – 1) (x + 1)
Ans. (i) p(y) = y2 – y + 1
∵ p(y) = y2 – y + 1 = (y)2 – y + 1
∴ p(0) = (0)2 – (0) + 1 = 0 – 0 + 1 = 1
p(1) = (1)2 – (1) + 1 = 1 – 1 + 1 = 1
p(2) = (2)2 – 2 + 1 = 4 – 2 + 1 = 3
(ii) p(t) = 2 + t + 2t2 – t3
∵ p(t) = 2 + t + 2t2 – t3 = 2 + t + 2(t)2 – (t)3
∴ p(0) = 2 + (0) + 2(0)2 – (0)3
= 2 + 0 + 0 – 0 = 2
p(1) = 2 + (1) + 2(1)2 – (1)3
= 2 + 1 + 2 – 1 = 4
p(2) = 2 + 2 + 2(2)2 – (2)3 = 2 + 2 + 8 – 8 = 4 (iii) p(x) = x3
∵ p(x) = x3 = (x)3
∴ p(0) = (0)3 = 0
p(1) = (1)3 = 1
p(2) = (2)3 = 8 [∵ 2 × 2 × 2 = 8]
(iv) p(x)= (x – 1)(x + 1)
∵ p(x) = (x – 1)(x + 1)
∴ p(0) = (0 – 1)(0 + 1) = –1 × 1 = –1
p(1) = (1 – 1)(1 + 1) = (0)(2) = 0
p(2) = (2 – 1)(2 + 1) = (1)(2) = 3
3. Verify whether the following are zeros of the polynomial, indicated against them.
Ans. (i) ∵ p(x) = 3x + 1
(iii) Since, p(x) = x2 – 1
∴ p(1) = (1)2 – 1 = 1 – 1 = 0
Since, p(1) = 0,
∴ x = 1 is a zero of x2 – 1.
Also p(–1) = (–1)2 – 1 = 1 – 1 = 0
i.e. p(–1) = 0,
∴ x = –1 is also a zero of x2 – 1.
(iv) We have p(x) = (x + 1)(x – 2)
∴ p(–1) = (–1 + 1)(–1 – 2) = (0)(–3) = 0
Since p(–1) = 0,
∴ x = –1 is a zero of (x + 1)(x – 1).
∴ Also, p(2) = (2 + 1)(2 – 2) = (3)(0) = 0
∴ Since p(2) = 0,
∴ x = 2 is also a zero of (x + 1)(x – 1).
(v) We have p(x) = x2
∴ p(0) = (0)2 = 0
Since p(0) = 0,
∴ 0 is a zero of x2.
(vi) We have p(x) = lx + m
(vii) We have p(x) = 3x2 – 1
(viii) We have p(x) = 2x + 1
is not a zero of 2x + 1.
4. Find the zero of the polynomial in each of the following cases:
(i) p(x) = x + 5 (ii) p(x) = x – 5 (iii) p(x) = 2x + 5 (iv) p(x) = 3x – 2 (v) p(x) = 3x (vi) p(x) = ax, a ≠ 0
(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.
Ans. (i) We have p(x) = x + 5
∴ p(x) = 0
⇒ x + 5 = 0
∴ or x = –5
∴ Thus, a zero of x + 5 is (–5).
(ii) We have p(x) = x – 5
∴ p(x) = 0
⇒ x – 5 = 0
∴ or x = 5
∴ Thus, a zero of x – 5 is 5.
(iii) We have p(x) = 2x + 5
∴ p(x) = 0
⇒ 2x + 5 = 0
or 2x = –5
Thus, a zero of 3x – 2 is
(v) Since p(x) = 3x
∴ p(x) = 0
⇒ 3x = 0
Thus, a zero of 3x is 0.
(vi) Since, p(x) = ax, a ≠ 0
∴ p(x) = 0
⇒ ax = 0
or
Thus, a zero of ax is