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When doing polynomial division can there be a remainder of x?

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Updated: 10/16/2024

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Q: When doing polynomial division can there be a remainder of x?

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What is a polynomial multiplication with a quotient of x 3 and a remainder of 2?

To get a quotient and a remainder, you would need to do a division, not a multiplication.

When dividing a polynomial F x by the binomial x - a a remainder not equal to zero tells you that x - a is not a of the polynomial?

factor

According to the Remainder theorem the remainder of the problem in which a polynomial F x is divided by the binomial x - a equals?

F(a)

When the polynomial in p(x) is divided by (x plus a)?

The result is a polynomial q(x) whose order is one fewer than the order of p(x) and a remainder term of the form b/(x + a).

What are the solutions of the equation x3 plus 3x2-x-3 equals 0?

the solutions to this equation are -1,+1 and -3. you can solve this equation by using the polynomial long division method. we basically want to factorize this and polynomial and equate its factors to zero and obtain the roots of the equation. By hit and trial , it clear that x=1 i.e is a root of this equation. So (x-1) should be a factor of the given polynomial (LHS). Divide the polynomial by x-1 using long division method and you will get the quotient as x2+4x+3 and remainder would be 0 ( it should be 0 as we are dividing the polynomial with its factor. Eg when 8 is divided by any of its factor like 4,2 .. remainder is always zero ) Now, we can write the given polynomial as product of its factors as x3+3x2-x-3 = (x-1)(x2+4x+3) =(x-1)(x+1)(x+3) [by splitting middle term method] so the solutions for the given polynomial are obtained when RHS = 0, Hence x=-1 , X = +1, x=-3 are the solutions for this equation.

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Related Questions

What is a polynomial multiplication with a quotient of x 3 and a remainder of 2?

To get a quotient and a remainder, you would need to do a division, not a multiplication.

What is polynomial division?

That means that you divide one polynomial by another polynomial. Basically, if you have polynomials "A" and "B", you look for a polynomial "C" and a remainder "R", such that: B x C + R = A ... such that the remainder has a lower degree than polynomial "B", the polynomial by which you are dividing. For example, if you divide by a polynomial of degree 3, the remainder must be of degree 2 or less.

When dividing the polynomial x3 4x2 5x 2 by x 2 the remainder is 0 making x 2 a factor?

Yes, if there is no remainder after division, the divisor is a factor.

If a polynomial is divided by (x-a) and the remainder equals zero then (x-a) is a factor of the polynomial?

false - apex

When dividing a polynomial F x by the binomial x - a a remainder not equal to zero tells you that x - a is not a of the polynomial?

factor

What do we use the polynomial remainder theorem for?

If a polynomial is divided by x - c, we can use the Remainder theorem to evaluate the polynomial at c.The Remainder theorem:If the polynomial f(x) is divided by x - c, then the remainder is f(c).Example:Given f(x) = x^3 - 4x^2 + 5x + 3, use the remainder theorem to find f(2).Solution:By the remainder theorem, if f(x) is divided by x - 2, then the remainder is f(2).We can use the synthetic division to divide.2] 1 -4 5 32 -4 2__________1 -2 1 5The remainder is 5, so f(2) = 5Check:f(x) = x^3 - 4x^2 + 5x + 3f(2) = (2)^3 - 4(2)^2 + 5(2) + 3 = 8 - 16 + 10 + 3 = 5

According to the Remainder theorem the remainder of the problem in which a polynomial F x is divided by the binomial x - a equals?

F(a)

What is divide x squared 3x-2byx-2?

To divide (x^2 + 3x - 2) by (x - 2), you can use polynomial long division or synthetic division. The result of dividing these two polynomials is (x + 5), with a remainder of 8.

Is it false if a polynomial is divided by (x-a) and the remainder equals zero then (x-a) is a factor of the polynomial?

No, it’s true. It’s the same as saying if 60 is divided by 2 and the remainder equals zero (no remainder, so it divides perfectly), 2 is a factor of 60.

When the polynomial in p(x) is divided by (x plus a)?

The result is a polynomial q(x) whose order is one fewer than the order of p(x) and a remainder term of the form b/(x + a).

When a certain polynomial is divided by x - 3 its quotient is xsquared - 5x - 6 and its remainder is 5 What is the polynomial?

From the Division Algorithm for Polynomials theorem,f(x) = q(x)g(x) + r(x) or we say:dividend = (quotient)(divisor) + (remainder)In our case,quotient = x^2 - 5x - 6; divisor = x - 3; and remainder = 5.Substitute what you know into the formula, and you will have:f(x) = (x^2 - 5x - 6)(x - 3) + 5f(x) = x^3 - 5x^2 - 6x - 3x^2 + 15x + 18 + 5f(x) = x^3 - 5x^2 - 3x^2 - 6x + 15x + 18 + 5f(x) = x^3 - 8x^2 + 9x + 23 (this is the required polynomial)

What are the solutions of the equation x3 plus 3x2-x-3 equals 0?

the solutions to this equation are -1,+1 and -3. you can solve this equation by using the polynomial long division method. we basically want to factorize this and polynomial and equate its factors to zero and obtain the roots of the equation. By hit and trial , it clear that x=1 i.e is a root of this equation. So (x-1) should be a factor of the given polynomial (LHS). Divide the polynomial by x-1 using long division method and you will get the quotient as x2+4x+3 and remainder would be 0 ( it should be 0 as we are dividing the polynomial with its factor. Eg when 8 is divided by any of its factor like 4,2 .. remainder is always zero ) Now, we can write the given polynomial as product of its factors as x3+3x2-x-3 = (x-1)(x2+4x+3) =(x-1)(x+1)(x+3) [by splitting middle term method] so the solutions for the given polynomial are obtained when RHS = 0, Hence x=-1 , X = +1, x=-3 are the solutions for this equation.

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