A polynomial leaves remainder 5 when divided by x−3 and remainder 7 when divided by x−4. If the polynomial is divided by (x−3) (x−4), find remainder?
We have the remainder theorem as, if a polynomial ƒ(x) is divided by (x-k), then the remainder is r=f(k).Now proceed as follows:Let, p(x) be the polynomial the question is referring to.Hence, according to given conditions, we can write p(x)/(x-3)=f(x)+5/(x-3) and p(x)/(x-4)=g(x)+7/(x-3) where f(x) and g(x) are the assumed quotient polynomials and 5 and 7 are the remainders as given.Hence, above two equations can be rearranged as,p(x)=(x-3)f(x)+5 and p(x)=(x-4)g(x)+7Now, according to remainder theorem, p(3)=5 and p(5)=7 . We know that, the degree of remainder polynomial resulted by taking the ratio of two polynomials is always less than the degree of the divisor polynomial by at least 1.As the divisor is (x-3)(x-4) i.e. a quadratic polynomial i.e. a polynomial of degree 2, hence the remainder polynomial has to be of the form of (ax+b) i.e. degree of remainder polynomial will be less than divisor polynomial by at least 1 . Consider a quotient polynomial h(x) which is obtained after dividing p(x) by (x-3)(x-4) and the remainder of this division operation will be of the form of (ax+b) . This can be written mathematically as, p(x)=(x-3)(x-4)h(x)+(ax+b)Putting the given conditions in this equation, we getp(3)=0+3a+b i.e. 3a+b=5 andp(4)=0+4a+b i.e. 4a+b=7Solving these two simultaneous equations, we get a=2, b=-1Hence, the remainder will be (2x-1)
A polynomial f(x) has a remainder 9 when divided by (x-3) and remainder -5 when divided by (2x+1), find...?
f(3) = 9, and f(-1/2) = -5. We want the remainder when f is divided by (x-3)(2x+1). So let f(x) = q(x)(x-3)(2x+1) + r(x), where q is the quotient and r is the remainder. Being the remainder on division by a quadratic, r must be a linear polynomial, say r = ax+b. So f(x) = q(x)(x-3)(2x+1) + ax + b f(3) = 9 means 3a+b = 9. f(-1/2) = -5 means -1/2a + b = -5, or -a + 2b = -10 Solve the system (I'll multiply the second equation by 3 to eliminate a): 3a+b=9, -3a+6b=-30. Adding tells us that 7b= -21, and b = -3. Substitute that into 3a+b=9, so 3a=12, a=4. Conclusion: the remainder is ax+b = 4x-3.
The polynomial f(x) X4 -2x3+3x2-ax+b when divided by (x-1) and (x+1) leaves the remainders 5 and 19. What is the value of a and b and what is the remainder when f(x) is divided by (x-3)?
f(x) x^4–2x^3+3x^2-ax+bPut x = 1 to get, 1–2+3-a+b = 0, or -a+b=-2 …(1)Put x = -1 to get, 1+2+3+a+b = 0, or a+b = -6 …(2)Add (1) and (2) to get: 2b = -8, or b = -4. from (2), a = -6+4 = -2.So the polynomial is x^4–2x^3+3x^2+2x-4 = 0.Check: x^4–2x^3+3x^2+2x-4 with x=1, becomes 1–2+3+2–4 = 0 = RHS. Correctx^4–2x^3+3x^2+2x-4 with x=11, becomes 1+2+3-2–4 = 0 = RHS. Correctx^4–2x^3+3x^2+2x-4 is divided by (x-3), the remainder will be 81–54+27+6–4 = 56
The polynomial p(x) =x^4-2x^3+3x^2-ax+b when divided by (x-1) and (x+1) leaves the remainders 5 and 19, respectively. What are a and b?
We have,p(x)=x^4–2x^3+3x^2-ax+bBy remainder theorem, when p(x) is divided by (x-1) and (x+1) , the remainders are equal to p(1) and p(-1) respectively.By the given condition, we havep(1)=5 and p(-1)=19=> (1)^4–2(1)^3+3(1)^2-a(1)+b=5 and (-1)^4–2(-1)^3+3(-1)^2-a(-1)+b=19=> 1–2+3-a+b=5 and 1-(-2)+3+a+b=19=> -a+b=5–1+2–3 and 1+2+3+a+b=19=> -a+b=3 and a+b=19–1–2–3=> -a+b=3 and a+b=13Adding these two equations,we get-a+b+a+b=3+13=> 2b=16=> 2b/2=16/2=> b=8Putting b=8 in a+b=13 , we geta+8=13=> a=13–8=> a=5Therefore, a=5 and b=8 .
Find the quotient Q(x) and remainder R(x) when the polynomial P(x) is divided by the polynomial D(x).?
Can someone help me find the quotient for these two problems ? 1. P(x) = 3x4 + 2x3 − 5x + 2; D(x) = x2 + 2x − 1 2. P(x) = 3x5 + 4x4 − 3x3 + x2 + x − 1; D(x) = x4 + x3 − 3x − 1
What is the remainder when -2 is synthetically divided into the polynomial 2x3 - x2+3x-4?!?
2 2 -1 3 -4 4 6 18 ---------------------------------- 2 3 9 14 14
When (x^4)-(4x^3)-(ax^2) + bx + 1 is divided by (x-1), the remainder is 7. When it is divided by (x+1), the...?
When (x^4)-(4x^3)-(ax^2) + bx + 1 is divided by (x-1), the remainder is 7. When it is divided by (x+1), the remainder is 3. Determine the values of a and b. Please provide a solution so I can learn ! + 10 points for best answer!
When a polynomial f(x) is divided by (x-1) and (x-2), it leaves remainder 5 and 7 respectively. What is the remainder when divided by (x-1) (x-2)?
F(x) = (x-1) q(x) + 5 …..1F(x) = (x-2) q'(x) + 7 ……2So F(1) = 5 and F(2) = 7.Now F(x) is divided by (x-1)(x-2) it's is a polynomial of degree two.Reminder should be a polynomial of degree less then 2. Say.. r=ax+bNow.. F(x) = (x-1)(x-2) q”(x) + rF(x) = (x-1)(x-2) q”(x) + (ax+b). ………3@x=1. 5= a+b@x=2. 7= 2a+bSolve two equations and get the value of a and b. That's the ans.r = 2x+3.