Mathematics, 03.08.2019 23:30 tinacalderon2856

Not really a question but..someone explain to me how to divide polynomials! i can’t get it

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This is division: $\dfrac{\text{dividend}}{\text{divisor}} = \text{quotient} + \dfrac{\text{remainder}}{\text{divisor}}$ for example, 5 divided by 2 is 2 with a remainder of 1. so the division statement is $\dfrac{5}{2} = 2 + \dfrac{1}{2}$ if you did long division, you'd repeatedly subtract after multiplying 2 by some number, etc.polynomial long divisionsee the attachment headed with "polynomial long division" for examplevery similar to the normal long division with numbers.if we want to divide $\left(3x^2+2x+7\right) \div (x+2)$ then we can first set up the division symbols. the dividend, $3x^2+2x+7$ , goes inside and the divisor, x+2, goes outside.when we do polynomial long division, we want to get rid of the highest x terms of the divisor. we can multiply $x+2$ by $3x$ to get $3x^2 + 6x$ , and then subtract that from the dividend to get rid of the $3x^2$ term. then we repeat until we get something that is of a lesser degree than the divisor. it is very similar to long division with whole numbers.in the example, 15 is a polynomial with a degree of 0; the dividend is of a degree of 1. we stop there and that's our remainder.the quotient is $3x-4$ and remainder is 15.the division statement would look like $\dfrac{3x^2+2x+7}{x+2} = 3x - 4 + \dfrac{15}{x+2}$ ======synthetic divisionsee attachment headed "synthetic division" for exampleread it. when we get to the final step, we get this set of numbers: $4 \quad -8 \quad 25 \quad -62$ starting from the right: -62 is the remainder 25 is the coefficient on the $x^0$ term, which is the constant -8 is the coefficient on the $x^1$ term 4 is the coefficient on the $x^2$ termso the quotient is $4x^2 - 8x + 25$ with a remainder of $-62$ this example represented in quotient form is $\dfrac{4x^3+9x-12}{x+2}=4x^2-8x+25+\dfrac{-62}{x+2}$ which is $\dfrac{4x^3+9x-12}{x+2}=4x^2-8x+25-\dfrac{62}{x+2}$

Answer from: Quest

may i see the picture? it would a lot, xox

Answer from: Quest

guess at question:

a vertical line intersects the graph of g(x,y) = 0 at two points. what conclusion follows about the function f(x) implicitly defined by g(x,f(x)) = 0?

the domain of f(x) cannot include the x coordinate of any such vertical line.

informally, f(x) is not a function, and

g(x,y)=0 cannot be solved for y uniquely.

the relation r = {(x,y) such that g(x,y) = 0} is a function if

for all (a,b) ∊ r and (c,d) ∊ r, a=c implies b=d.

step-by-step explanation:

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Not really a question but..someone explain to me how to divide polynomials! i can’t get it...

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