Question

(iii) প্ৰতিটো ৰে
(iv) এটা বহুপদ শূন্য এটাতকৈ বেU
অনুশীলনী 2.2
5r – 4x + 3 বহুপদৰ মান নির্ণয় কৰা যেতিয়া,
(i) x = 0 (ii) r = -1 (iii) 1 = 2
পদবােৰৰ প্রত্যেকৰ বাবে p(0), p) তা
(ii) p(t) -
(iv) p(x)
(i) p(v) = ] - ] + 1
(iii) p(x) = r।
কাষত উল্লেখিত মানবােৰ বহুপদটোৰ শূন্য হয়নে নত​

Answer 1

SOLUTION

TO DETERMINE

1. Find the value of the polynomial 5x - 4x² + 3 at

(i) x = 0 (ii) x = - 1 (iii) x = 2

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 )

EVALUATION

ANSWER TO QUESTION : 1

Let p(x) be the given polynomial then

$\sf{p(x) = 5x - 4 {x}^{2} + 3}$

(i) Putting x = 0 we get

$\sf{p(0) = 5.0 - 4 .{0}^{2} + 3 = 0 - 0 + 3 = 3}$

(ii) Putting x = - 1 we get

$\sf{p( - 1) = 5.( - 1) - 4 .{( - 1)}^{2} + 3 = - 5 - 4 + 3 = - 6}$

(iii) Putting x = 2 we get

$\sf{p(2) = 5.( 2) - 4 .{( 2)}^{2} + 3 = 10 - 16 + 3 = - 3}$

ANSWER TO QUESTION : 2

(i) Here the given polynomial is

$\sf{p(y) = {y}^{2} - y + 1 }$

Thus we get

$\sf{p(0) = {0}^{2} - 0+ 1 = 0 - 0 + 1 = 1 }$

$\sf{p(1) = {1}^{2} - 1+ 1 = 1 - 1 + 1 = 1 }$

$\sf{p(2) = {2}^{2} - 2+ 1 = 4 - 2 + 1 = 3}$

(ii) Here the given polynomial is

$\sf{p(t) = 2 + t + 2 {t}^{2} - {t}^{3} }$

Thus we get

$\sf{p(0) = 2 + 0 + 2 .{0}^{2} - {0}^{3} = 2 }$

$\sf{p(1) = 2 + 1 + 2 .{1}^{2} - {1}^{3} = 4 }$

$\sf{p(2) = 2 + 2 + 2 .{2}^{2} - {2}^{3} = 4 }$

(iii) Here the given polynomial is

$\sf{p(x) = {x}^{3} }$

Thus we get

$\sf{p(0) = {0}^{3} = 0 }$

$\sf{p(1) = {1}^{3} = 1}$

$\sf{p(2) = {2}^{3} = 8}$

(iv) Here the given polynomial is

$\sf{p(x) = (x - 1)(x + 1)}$

Thus we get

$\sf{p(0) = (0 - 1)(0 + 1) = - 1}$

$\sf{p(1) = (1 - 1)(1 + 1) = 0 \times 2 = 0}$

$\sf{p(2) = (2- 1)(2+ 1) = 1 \times 3 = 3}$

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