VERY SHORT ANSWER TYPE QUESTIONS
Q1. Write an example of an algebraic expression that is not a polynomial.
Q2. p(x) = √
3 + 1 is not a polynomial. Give reason
Q3. Find the value of polynomial 8x3
- 6x2 +2 at x = 1
Q4. If p(x) = 6x3 + 5x2 – 3x + 2 find p(-1)
Q5. Find the zero of the polynomial p(y) = 2y + 7
Q6. Find the remainder when x101 – 1 is divided by x -1
Q7. Find whether xn + yn is divisible by x – y ( y ≠ 0) or not.
Q8. Write the following polynomials in standard form
i.4y- 4y3 +3 –y
4
ii.5m3
-6m +7 -2m2
Q9. Write the integral zeroes of the following polynomials
i.( x – 3) ( x – 7 )
ii.( x + 1 ) ( 3x + 2 )
pls answer it is important tomorrow is my exam
Answers
Answer:
Q1. Ans- For example the expressions √x=x12,x−3=1x3 are algebraic expressions but they are not polynomials because the exponents 12,−3 on the variable x are fractions and negative integers.
Q2. Ans- p(x) = √
3 + 1 is not a polynomial because polynomial will never in underroot form.
Q3. Ans- x+1 = 0
x = -1
8(-1)*3-6-(-1)*2+2
= -8-6+2
= -14+2
= -12
Q4.Ans- If p = (-1),
Then,
= -6 + 5 + 3 +2
= -6 + 10
= 4
Q5. Ans- Zero of 2y+7=>
2y+7=0
2y =-7
y=-7/2
Q6. Ans- What is the remainder when x^101 is divided by (x-1)? Therefore the remainder is 1.
Q7. Ans- Let P(n) : xn – yn is divisible by x – y, where x and y are any integers with x≠y.
Now, P(l): x1 -y1 = x-y, which is divisible by (x-y)
Hence, P(l) is true.
Let us assume that, P(n) is true for some natural number n = k.
P(k): xk -yk is divisible by (x – y)
or xk-yk = m(x-y),m ∈ N …(i)
Now, we have to prove that P(k + 1) is true.
P(k+l):xk+l-yk+l
= xk-x-xk-y + xk-y-yky
= xk(x-y) +y(xk-yk)
= xk(x – y) + ym(x – y) (using (i))
= (x -y) [xk+ym], which is divisible by (x-y)
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.
Q8. i. Ans- -y4-4y3+4y+3
ii. Ans- 5m3-2m2-6m+7
Q9. i. (x-3)(x-7)
x²-7x-3x+21
x²-10x+21
ii.(X+1)(3x+2)
3x²+2x+3x+2
3x²+5x+2
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