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CBSE - Class 9 Lessons
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Post a LessonAnswered 3 days ago Learn CBSE - Class 9/Social Studies/History/Socialism in Europe and the Russian Revolution
Sana Begum
I am online Quran teacher 7 years
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 02 Sep Learn CBSE - Class 9/Class 9 Mathematics/Chapter 2 Polynomials/EXERCISE - 2.4
Math Decode Institute
i)
(ii)
(iii)
Answered on 22 Aug Learn CBSE - Class 9/Class 9 Mathematics/Chapter 1 Number Systems/EXERCISE - 1.5
Shivam Yadav
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