x class polynomials exercise-1
EXERCISE–1
Do this
1. State which of the following are polynomials and which are not? Give reasons.
(i) 2x3(ii)`1/(x–1)`(iii) 4z2+`1/7`(iv) `m^2– sqrt2 m +2` (v) p–2
Solution:(i) 2x3 Is a polynomial.Because index of x is a non–negative integer.
(ii)`1/(x–1)`Is not a polynomial. Because index of x is a negative integer.
(iii) 4z2 Is a polynomial.Because index of z is a non–negative integer.
(iv) `m^2– sqrt2 m +2` Is a polynomial.Because index of m is a non–negative integer.
(v) p–2 Is not a polynomial. Because index of p is a negative integer.
2.p(x) = x2 – 5x – 6, find the values of p(1), p(2), p(3), p(0), p(–1), p(–2), p(–3).
Solution:p(1) = (1)2 – 5(1) – 6 = 1 – 5 – 6 = – 10
p(2) = (2)2 – 5(2) – 6 = 4 – 10 – 6 = –12
p(3) = (3)2 – 5(3) – 6 = 9 – 15 – 6 =9 –21 = – 12
p(0) = (0)2 – 5(0) – 6 =0– 0–6 = – 6
p(–1) = (–1)2 – 5(–1) – 6 = 1 + 5 – 6 = 0
p(–2) = (–2)2 – 5(–2) – 6 = 4 + 10 – 6 = 14 – 6 =8
p(–3) = (–3)2 – 5(–3) – 6 = 9 + 15 – 6 = 24 – 6 =18
3.p(m) = m2 – 3m + 1, find the value of p(1) and p(–1).
Solution:p(1) = (1)2 – 3(1) +1 = 1 – 3 +1 = – 1
p(–1) = (–1)2 – 3(–1) +1 = 1 + 3 +1 = 5
4. Let p(x) = x2 – 4x + 3. Find the value of p(0), p(1), p(2), p(3) and obtain zeroes of the polynomial p(x).
Solution:p(1) = (1)2 – 4(1) + 3 = 1 – 4 + 3 = 0
p(2) = (2)2 – 4(2) + 3 = 4 – 8 + 3 = 7 – 8 = –1
p(3) = (3)2 – 4(3) + 3 = 9 – 12 + 3 = 12 – 12 = 0
So, the zeroes of polinomial p(x) is 1 and 3
5. Check whether –3 and 3 are the zeroes of the polynomial x2 – 9.
Solution:p(–3) = (–3)2– 9 = 9 – 9 = 0
p(3) = 3 2 – 9 = 9 – 9 = 0
p(3) = 0 and p(–3) = 0 So, –3 and 3 are the zeroes of the polynomial x2 – 9.
Exercise:
1. (a) If p(x) = 5x7 – 6x5 + 7x–6, find (i) coefficient of x5 (ii) degree of p(x) (iii) constant term.
Solution:(i) coefficient of x5 = – 6
(ii) degree of p(x) = 7
(iii) constant term = – 6
2. State which of the following statements are true and which are false? Give reasons for your choice.
(i) The degree of the polynomial `sqrt2`x2– 3x + 1 is `sqrt2`.
False. `sqrt2` is the coefficient of x. The degree of the polynomial is 2.
(ii) The coefficient of x2 in the polynomial p(x) = 3x3 – 4x2 + 5x + 7 is 2.
False. The coefficient of x2 in the polynomial p(x) is –4
(iii) The degree of a constant term is zero.
True. There is no variable in costant term.
(iv) `1/(x^2 – 5x +6 )` is a quadratic polynomial.
False. This is not a polynomial because index of x is a negative integer.
(v) The degree of a polynomial is one more than the number of terms in it.
False. The terms of a Polynomial in general form is one more than the degree of a polynomial.
3. If p(t) = t3 – 1, find the values of p(1), p(–1), p(0), p(2), p(–2).
Solution:p(1) = 13 – 1 = 1 – 1 = 0
p(–1) = (–1)3 – 1 = –1 –1 = –2
p(0) = 03 – 1 = 0 –1 = –1
p(2) = 23 – 1 = 8 – 1 = 7
p(–2) = (–2)3 – 1 = –8 – 1 = – 9
4. Check whether –2 and 2 are the zeroes of the polynomial x4 – 16
Solution:p(–2) = (–2)4 – 16 = 16 – 16 = 0
p(2) = 24 – 16 = 16 – 16 = 0
p(2) = 0 and p(–2) = 0 So, –2 and 2 are the zeroes of the polynomial x2 – 16.
5. Check whether 3 and –2 are the zeroes of the polynomial p(x) when p(x) = x2 – x – 6.
Solution:p(3) = 32 – 3 – 6 = 9 – 9 = 0
p(–2) = (–2)2 – (–2) – 6 = 4 + 2 – 6 = 0
So, –2 and 2 are the zeroes of the polynomial p(x) = x2 – x – 6.
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