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Let a = 0 and p(x) be a polynomial of degree greater than 2. If p(x) leaves remainders a and – a when divided
respectively by x + a and x – a, the remainder when p(x) is divided by x2 – a2 is [EAMCET-2003]
- a)2x
- b)– 2x
- c)x
- d)– x
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
Let a = 0 and p(x) be a polynomial of degree greater than 2. If p(x) l...
Let the remainder be R(x), then
R(x) = p(x)+q and R(a) = -a
Given R(-a) = a pa+q = -a -----(2)
-pa+q = a ----(1)
Solving (1) & (2), we get
p=-1, q=0
Therefore R(x) = -x
Most Upvoted Answer
Let a = 0 and p(x) be a polynomial of degree greater than 2. If p(x) l...
A+1 when divided by x and x+1, respectively, then p(x) has at least three distinct roots.
Proof:
Since p(x) leaves a remainder of 0 when divided by x-a, we know that p(a) = 0. Similarly, since p(x) leaves a remainder of a+1 when divided by x-(a+1), we know that p(a+1) = a+1.
Suppose for contradiction that p(x) has only two roots, say r and s, where r ≠ s. Then p(x) can be factored as p(x) = c(x-r)(x-s), where c is a constant. Since p(x) has degree greater than 2, we know that c ≠ 0.
Since p(a) = 0, we have c(a-r)(a-s) = 0. Since c ≠ 0, we must have either a = r or a = s. Without loss of generality, assume that a = r. Then p(x) can be written as p(x) = c(x-a)(x-s) = cx^2 - c(a+s)x + cas.
Now, since p(a+1) = a+1, we have c(a+1)^2 - c(a+s)(a+1) + cas = a+1. Simplifying, we get c(s-a) = 1, which implies that c ≠ 0 and s ≠ a.
Now, consider the polynomial q(x) = p(x) - (x-a)(x-(a+1)). Since p(x) and (x-a)(x-(a+1)) have the same remainders when divided by x and x+1, respectively, we have q(a) = q(a+1) = 0.
We can write q(x) as q(x) = cx^2 - c(a+s+1)x + cas + a(a+1). Note that q(x) has degree 2 and has roots a and a+1. Therefore, q(x) must be of the form q(x) = c(x-a)(x-(a+1)) = cx^2 - c(a+s+1)x + cas + a(a+1).
Comparing coefficients, we have c = 1 and cas + a(a+1) = 0. Since s ≠ a, we have c ≠ 0, which implies that s ≠ -(a+1). Therefore, we have cas + a(a+1) ≠ 0, which is a contradiction.
Therefore, our assumption that p(x) has only two roots is false. Thus, p(x) must have at least three distinct roots.
Proof:
Since p(x) leaves a remainder of 0 when divided by x-a, we know that p(a) = 0. Similarly, since p(x) leaves a remainder of a+1 when divided by x-(a+1), we know that p(a+1) = a+1.
Suppose for contradiction that p(x) has only two roots, say r and s, where r ≠ s. Then p(x) can be factored as p(x) = c(x-r)(x-s), where c is a constant. Since p(x) has degree greater than 2, we know that c ≠ 0.
Since p(a) = 0, we have c(a-r)(a-s) = 0. Since c ≠ 0, we must have either a = r or a = s. Without loss of generality, assume that a = r. Then p(x) can be written as p(x) = c(x-a)(x-s) = cx^2 - c(a+s)x + cas.
Now, since p(a+1) = a+1, we have c(a+1)^2 - c(a+s)(a+1) + cas = a+1. Simplifying, we get c(s-a) = 1, which implies that c ≠ 0 and s ≠ a.
Now, consider the polynomial q(x) = p(x) - (x-a)(x-(a+1)). Since p(x) and (x-a)(x-(a+1)) have the same remainders when divided by x and x+1, respectively, we have q(a) = q(a+1) = 0.
We can write q(x) as q(x) = cx^2 - c(a+s+1)x + cas + a(a+1). Note that q(x) has degree 2 and has roots a and a+1. Therefore, q(x) must be of the form q(x) = c(x-a)(x-(a+1)) = cx^2 - c(a+s+1)x + cas + a(a+1).
Comparing coefficients, we have c = 1 and cas + a(a+1) = 0. Since s ≠ a, we have c ≠ 0, which implies that s ≠ -(a+1). Therefore, we have cas + a(a+1) ≠ 0, which is a contradiction.
Therefore, our assumption that p(x) has only two roots is false. Thus, p(x) must have at least three distinct roots.
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Community Answer
Let a = 0 and p(x) be a polynomial of degree greater than 2. If p(x) l...
Let P(x)=(x−a)(x+a)Q(x)+rx+s
When we divide by x+a, we get
P(−a)=(−a−a)(−a+a)Q(−a)+r(−a)+s=a
⇒s−ra=a ...(1)
And when we divide by x−a, we get
P(a)=(a−a)(a+a)Q(a)+ra+s=−a
⇒s+ra=−a ...(2)
Solving (1) and (2), we get
s=0,r=−1
Hence, P(x)=(x−a)(x+a)Q(x)−x
And when we divide by x2-a2, it leaves a remainder −x.
Hence option D is correct
When we divide by x+a, we get
P(−a)=(−a−a)(−a+a)Q(−a)+r(−a)+s=a
⇒s−ra=a ...(1)
And when we divide by x−a, we get
P(a)=(a−a)(a+a)Q(a)+ra+s=−a
⇒s+ra=−a ...(2)
Solving (1) and (2), we get
s=0,r=−1
Hence, P(x)=(x−a)(x+a)Q(x)−x
And when we divide by x2-a2, it leaves a remainder −x.
Hence option D is correct
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Let a = 0 and p(x) be a polynomial of degree greater than 2. If p(x) leaves remainders a and – a when dividedrespectively by x + a and x – a, the remainder when p(x) is divided by x2– a2is [EAMCET-2003]a)2xb)– 2xc)xd)– xCorrect answer is option 'D'. Can you explain this answer?
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Let a = 0 and p(x) be a polynomial of degree greater than 2. If p(x) leaves remainders a and – a when dividedrespectively by x + a and x – a, the remainder when p(x) is divided by x2– a2is [EAMCET-2003]a)2xb)– 2xc)xd)– xCorrect answer is option 'D'. Can you explain this answer? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about Let a = 0 and p(x) be a polynomial of degree greater than 2. If p(x) leaves remainders a and – a when dividedrespectively by x + a and x – a, the remainder when p(x) is divided by x2– a2is [EAMCET-2003]a)2xb)– 2xc)xd)– xCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let a = 0 and p(x) be a polynomial of degree greater than 2. If p(x) leaves remainders a and – a when dividedrespectively by x + a and x – a, the remainder when p(x) is divided by x2– a2is [EAMCET-2003]a)2xb)– 2xc)xd)– xCorrect answer is option 'D'. Can you explain this answer?.
Let a = 0 and p(x) be a polynomial of degree greater than 2. If p(x) leaves remainders a and – a when dividedrespectively by x + a and x – a, the remainder when p(x) is divided by x2– a2is [EAMCET-2003]a)2xb)– 2xc)xd)– xCorrect answer is option 'D'. Can you explain this answer? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about Let a = 0 and p(x) be a polynomial of degree greater than 2. If p(x) leaves remainders a and – a when dividedrespectively by x + a and x – a, the remainder when p(x) is divided by x2– a2is [EAMCET-2003]a)2xb)– 2xc)xd)– xCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let a = 0 and p(x) be a polynomial of degree greater than 2. If p(x) leaves remainders a and – a when dividedrespectively by x + a and x – a, the remainder when p(x) is divided by x2– a2is [EAMCET-2003]a)2xb)– 2xc)xd)– xCorrect answer is option 'D'. Can you explain this answer?.
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