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Post a LessonAnswered on 02 Sep Learn CBSE - Class 9/Class 9 Mathematics/Chapter 2 Polynomials/EXERCISE - 2.4
Math Decode Institute
i)
(ii)
(iii)
Answered on 09 Aug Learn CBSE - Class 9/Class 9 Mathematics/Chapter 2 Polynomials/EXERCISE - 2.2
Rakhi Yadav
more than 5 year experience tutor
Okay, let's find the values of the polynomials at the specified points.
(i) p(y) = y2 – y + 1
p(0) = 02 – 0 + 1 = 1
p(1) = 12 – 1 + 1 = 1
p(2) = 22 – 2 + 1 = 3
(ii) p(t) = 2 + t + 2t2 – t3
p(0) = 2 + 0 + 2(0)2 – 03 = 2
p(1) = 2 + 1 + 2(1)2 – 13 = 4
p(2) = 2 + 2 + 2(2)2 – 23 = 4
(iii) p(x) = x3
p(0) = 03 = 0
p(1) = 13 = 1
p(2) = 23 = 8
(iv) p(x) = (x – 1) (x + 1)
p(0) = (0 – 1) (0 + 1) = -1
p(1) = (1 – 1) (1 + 1) = 0
p(2) = (2 – 1) (2 + 1) = 3
Here are the final answers:
(i) p(0) = 1, p(1) = 1, p(2) = 3
(ii) p(0) = 2, p(1) = 4, p(2) = 4
(iii) p(0) = 0, p(1) = 1, p(2) = 8
(iv)
p(0) = -1, p(1) = 0, p(2) = 3
read lessAnswered on 11 Aug Learn CBSE - Class 9/Class 9 Mathematics/Chapter 2 Polynomials/EXERCISE - 2.3
Sanjeev Verma
Answered on 31 Jul Learn CBSE - Class 9/Class 9 Mathematics/Chapter 2 Polynomials/EXERCISE - 2.3
Satyansh Singh
Answered on 11 Aug Learn CBSE - Class 9/Class 9 Mathematics/Chapter 2 Polynomials/EXERCISE - 2.4
Sanjeev Verma
in this question we will use the identity
(ax + b) * (cx + d) = ac*x^2 + (ad + bc)* x + bd
(1) (x + 4) (x + 10) = x^2 + (4 + 10)*x + 4 * 10
=x^2 + 14x + 40
(2) (x +8) (x - 10) = x^2 + (8 - 10)x + 8 * -10
= x^2 - 2x - 80
(3)(3x + 4) (3x - 5) = (3x)^2 + (4*3 + (- 5)*3) * x + 4 * (-5)
= 9x^2 - 3x - 20
read lessAsk a Question
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