Degree of p(x) = degree of q(x) The degree of dividend and quotient can be equal, if the divisor is a constant (degree 0) term. Therefore Let p(x) = 3x - 6x + 5 Let g(x) = 3 Therefore, q(x) = x² - 2x + 1 and r(x) = 2 Video Solution 👀👇 Euclid's division lemma can also be generalized to polynomials. FRead more
Degree of p(x) = degree of q(x)
The degree of dividend and quotient can be equal, if the divisor is a constant (degree 0) term. Therefore
Let p(x) = 3x – 6x + 5
Let g(x) = 3
Therefore, q(x) = x² – 2x + 1 and r(x) = 2
Video Solution 👀👇
Euclid’s division lemma can also be generalized to polynomials. For example, suppose we want to divide the polynomial x³ + 2x² – 3x + 4 by the polynomial x – 1. According to Euclid’s division lemma, there exist unique polynomials Q and R such that x³ + 2x² – 3x + 4 = (x – 1) Q + R, where the degree of R is less than the degree of x – 1. In this case, we can see that Q = x² + 3x – 4 and R = -x + 4 satisfy the conditions of the lemma, so the quotient is x² + 3x – 4 and the remainder is -x + 4 when x³ + 2x² – 3x + 4 is divided by x – 1.
Find: (i) 64^(1/2) (ii) 32^(1/5) (iii) 125^(1/3)
(i) 64^(1/2) = (8^2)^1/2 = 8 (ii) 32^1/2 = (2^5)^1/5 = 2^(5×1/5) = 2 (iii) 125^1/3 = (5^3)^1/3 = 5^(3×1/3) = 5
(i) 64^(1/2) = (8^2)^1/2 = 8
See less(ii) 32^1/2 = (2^5)^1/5 = 2^(5×1/5) = 2
(iii) 125^1/3 = (5^3)^1/3 = 5^(3×1/3) = 5
Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer. 4x² – 3x + 7
4x² – 3x + 7 Polynomials in one variable as it contains only one variable x.
4x² – 3x + 7 Polynomials in one variable as it contains only one variable x.
See lessSimplify: (i) 2^(2/3). 2^(1/5) (ii) (1/3^3)^7 (iii) (11^1/2)/(11^1/4) (iv) 7^(1/2) . 8^(1/2)
(i) 2^(2/3).2^(1/5) = 2^((2/3)+(1/5)) = 2^((10+3)/15) = 2^(13/15) (ii) (1/3^3)^7 = (3^-3)^7 = 3^-21 (iii) (11^1/2)/(11^1/4) = 11^1/2 × 11^-1/4 = 11^((1/2)-(1/4)) = 11^(2-1 / 4) = 11^1/4 (iv) 7^1/2 . 8^1/2 = (7 × 8)^1/2 = 56^1/2
(i) 2^(2/3).2^(1/5) = 2^((2/3)+(1/5)) = 2^((10+3)/15) = 2^(13/15)
See less(ii) (1/3^3)^7 = (3^-3)^7 = 3^-21
(iii) (11^1/2)/(11^1/4) = 11^1/2 × 11^-1/4 = 11^((1/2)-(1/4)) = 11^(2-1 / 4) = 11^1/4
(iv) 7^1/2 . 8^1/2 = (7 × 8)^1/2 = 56^1/2
Find : (i) 9^(3/2) (ii)32^(2/5) (iii) 16^(3/4) (iv) 125^(-1/3)
(i) 9^(3/2) = (3^2)^(3/2) = 3^(2×3/2) = 3^2 = 9 (ii) 32^(2/5) = (2^5)^(2/5) = 2^(5×2/5) = 2^2 = 4 (iii) 16^(3/4) = (2^4)^(3/2) = 2^(4×3/4) = 2^3 = 8 (iv) 125^(-1/3) = (5^3^(-1/3) = 5^(3×-1/3) = 5^-1 = 1/5 = 5
(i) 9^(3/2) = (3^2)^(3/2) = 3^(2×3/2) = 3^2 = 9
See less(ii) 32^(2/5) = (2^5)^(2/5) = 2^(5×2/5) = 2^2 = 4
(iii) 16^(3/4) = (2^4)^(3/2) = 2^(4×3/4) = 2^3 = 8
(iv) 125^(-1/3) = (5^3^(-1/3) = 5^(3×-1/3) = 5^-1 = 1/5 = 5
Obtain all other zeroes of 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeroes are √(5/3) and – √(5/3).
It is an important Question. See here 👀👇
It is an important Question.
See lessSee here 👀👇
On dividing x² – 3 x² + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and –2x + 4, respectively. Find g(x).
This Question is from Chapter 2 Exercise 2.3. Video Explanation 👀
This Question is from Chapter 2 Exercise 2.3.
See lessVideo Explanation 👀
Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and deg p(x) = deg q(x)
Degree of p(x) = degree of q(x) The degree of dividend and quotient can be equal, if the divisor is a constant (degree 0) term. Therefore Let p(x) = 3x - 6x + 5 Let g(x) = 3 Therefore, q(x) = x² - 2x + 1 and r(x) = 2 Video Solution 👀👇 Euclid's division lemma can also be generalized to polynomials. FRead more
Degree of p(x) = degree of q(x)
The degree of dividend and quotient can be equal, if the divisor is a constant (degree 0) term. Therefore
Let p(x) = 3x – 6x + 5
Let g(x) = 3
Therefore, q(x) = x² – 2x + 1 and r(x) = 2
Video Solution 👀👇
See lessEuclid’s division lemma can also be generalized to polynomials. For example, suppose we want to divide the polynomial x³ + 2x² – 3x + 4 by the polynomial x – 1. According to Euclid’s division lemma, there exist unique polynomials Q and R such that x³ + 2x² – 3x + 4 = (x – 1) Q + R, where the degree of R is less than the degree of x – 1. In this case, we can see that Q = x² + 3x – 4 and R = -x + 4 satisfy the conditions of the lemma, so the quotient is x² + 3x – 4 and the remainder is -x + 4 when x³ + 2x² – 3x + 4 is divided by x – 1.
Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm And deg q(x) = deg r(x)
Degree of q(x) = degree of r(x) Let p(x) = 2x² - 4x + 3 Let g(x) =x² - 2x + 1 Therefore, q(x) = 2 and r(x) = 1 Explanation 😃👇
Degree of q(x) = degree of r(x)
Let p(x) = 2x² – 4x + 3
Let g(x) =x² – 2x + 1
Therefore, q(x) = 2 and r(x) = 1
Explanation 😃👇
See lessGive examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm And deg r(x) = 0
degree of r(x) = 0 Let p(x) = 2x² - 4x + 3 Let g(x) = x - 2x + 1 Therefore, q(x) = 2 and r(x) = 1 See here =>
degree of r(x) = 0
Let p(x) = 2x² – 4x + 3
Let g(x) = x – 2x + 1
Therefore, q(x) = 2 and r(x) = 1
See here =>
See lessCheck whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial: t²– 3, 2t⁴ + 3t³ – 2t² – 9t – 12
Exercise 2.3 Question no. 2. See Here ✌
Exercise 2.3 Question no. 2. See Here ✌
See less