Maths c4 OCR - diving polynomials...How to use identity method?

this be the question i'm tryna solve.

so obviously there's the long division method, but i've tried many times and i've never understood it. in c2 i just used the identity method for polynomials

we weren't taught the identity method for diving polynomials in class, but the mark scheme used that method as an example of how to solve this question

so im wondering if someone can explain to me how to use the identity method for these types of questions?

thank you

In the exam you can use any method you want as they never usually specify. So I'd say get good with whichever method you prefer and perfect that.

In this case for using the identity method, the key thing is to notice that the polynomial on the numerator is 1 degree higher than the denominator (you got x cubed on top and x squared on bottom). So you can write the quotient as Ax + B + (Cx+D)/(x^2 +4x +1). If the numerator was say of degree 4, then you could write the quotient as Ax^2 + Bx + C + (Dx+E)/(x^2 +4x +1) etc

So you get (x^3 +2x^2 -6x -5)/(x^2 +4x +1) = Ax + B + (Cx+D)/(x^2 +4x +1) .

You then multiply both sides by (x^2 +4x +1) and begin to equate coefficients on the RHS and LHS to find the values of the constants A, B, C and D. Your remainder will be Cx+D.

thank you very much
!!!

i shall try and apply this method when i do some practise q's
you've been v.v helpful to me )

To generalise what he has said to make it clear for all questions, suppose we have two polynomials $f_m(x)$ and $g_n(x)$ of degrees $m,n$ respectively where $m \geq n$.

Then we have

$\dfrac{f_m(x)}{g_n(x)} = q_{m-n}(x) + \dfrac{r_k(x)}{g_n(x)}$

where $q_{m-n}(x)$ is the quotient polynomial of degree $m-n$, and $r_k(x)$ is the remainder polynomial of degree $k < n$.

The highest $k$ can be is $n-1$ therefore we set $g_k(x)$ as a polynomial of degree $n-1$ with undetermined coefficients which can determine later and see whether the degree reduces or not.

Example:

$\dfrac{x^5+x+1}{x^2+4x+1}$ has the numerator with degree 5 and denominator with degree 2, hence our quotient will have degree 3 hence we gonna have $Ax^3+Bx^2+Cx+D$. Then the remainder will have degree 1 at most hence it will be $Ex+F$.

So the form is $\dfrac{x^5+x+1}{x^2+4x+1} = Ax^3+Bx^2+Cx+D + \dfrac{Ex+F}{x^2+4x+1}$

Indeed, solving it yields that $\dfrac{x^5+x+1}{x^2+4x+1} = x^3-4x^2+15x-56 + \dfrac{210x+57}{x^2+4x+1}$

To generalise what he has said to make it clear for all questions, suppose we have two polynomials $f_m(x)$ and $g_n(x)$ of degrees $m,n$ respectively where $m \geq n$.

Then we have

$\dfrac{f_m(x)}{g_n(x)} = q_{m-n}(x) + \dfrac{r_k(x)}{g_n(x)}$

where $q_{m-n}(x)$ is the quotient polynomial of degree $m-n$, and $r_k(x)$ is the remainder polynomial of degree $k < n$.

The highest $k$ can be is $n-1$ therefore we set $g_k(x)$ as a polynomial of degree $n-1$ with undetermined coefficients which can determine later and see whether the degree reduces or not.

Example:

$\dfrac{x^5+x+1}{x^2+4x+1}$ has the numerator with degree 5 and denominator with degree 2, hence our quotient will have degree 3 hence we gonna have $Ax^3+Bx^2+Cx+D$. Then the remainder will have degree 1 at most hence it will be $Ex+F$.

So the form is $\dfrac{x^5+x+1}{x^2+4x+1} = Ax^3+Bx^2+Cx+D + \dfrac{Ex+F}{x^2+4x+1}$

Indeed, solving it yields that $\dfrac{x^5+x+1}{x^2+4x+1} = x^3-4x^2+15x-56 + \dfrac{210x+57}{x^2+4x+1}$

To generalise what he has said to make it clear for all questions, suppose we have two polynomials $f_m(x)$ and $g_n(x)$ of degrees $m,n$ respectively where $m \geq n$.

Then we have

$\dfrac{f_m(x)}{g_n(x)} = q_{m-n}(x) + \dfrac{r_k(x)}{g_n(x)}$

where $q_{m-n}(x)$ is the quotient polynomial of degree $m-n$, and $r_k(x)$ is the remainder polynomial of degree $k < n$.

The highest $k$ can be is $n-1$ therefore we set $g_k(x)$ as a polynomial of degree $n-1$ with undetermined coefficients which can determine later and see whether the degree reduces or not.

Example:

$\dfrac{x^5+x+1}{x^2+4x+1}$ has the numerator with degree 5 and denominator with degree 2, hence our quotient will have degree 3 hence we gonna have $Ax^3+Bx^2+Cx+D$. Then the remainder will have degree 1 at most hence it will be $Ex+F$.

So the form is $\dfrac{x^5+x+1}{x^2+4x+1} = Ax^3+Bx^2+Cx+D + \dfrac{Ex+F}{x^2+4x+1}$

Indeed, solving it yields that $\dfrac{x^5+x+1}{x^2+4x+1} = x^3-4x^2+15x-56 + \dfrac{210x+57}{x^2+4x+1}$

oml i just saw this, thank you xxxx

bit trivial but, why do you use degree 3 instead of saying power 3?

think i'll stick with long division