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.
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.
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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.
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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 =>
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