#factorising? very confused

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half treeBOT
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charred rover
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Question 4🙏🙏

charred rover
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@charred rover

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.

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oki

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lets see

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ya

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ok so, since you know x^2-kx+1 is a factor

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$(x^2-kx+1)(px+q)=ax^3+bx^2+cx+d$

prime ospreyBOT
charred rover
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holds true

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how did you get px+q?

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p and q are just some values

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we will solve for them

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but as for now, they are unknown

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could i also say ax+c?

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yes, but i wouldnt use a and c as they are already used in the question and mean specific things

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technically i should write it like this

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ah ok

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$(x^2-kx+1)(px+q)=ax^3+dx^2+bx+c$

prime ospreyBOT
charred rover
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we know that d=0

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because in the question, the x^2 term is 0

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its what is called a "depressed cubic"

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now what you can do, is expand out (x^2-kx+1)(px+q)

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is there a reason for x^2 being d and not b

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then equate like terms

charred rover
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if we order our algebra like this, we wont need to rename variables at the end

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oh yes my mistake

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so, since d=0

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$(x^2-kx+1)(px+q)=ax^3+bx+c$

prime ospreyBOT
charred rover
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as the question asks

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we can expand out the left hand side

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$(px^3-pkx^2+px+qx^2-qkx+q)=ax^3+bx+c$

prime ospreyBOT
charred rover
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simplify

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and would u factor like terms?

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so x^2(q-pk)

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$px^3+(q-pk)x^2+(p+qk)x+q=ax^3+bx+c$

prime ospreyBOT
charred rover
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yup

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exactly

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and now we can see that for it to work

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p = a

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q-pk = 0

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p+qk = b

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q = c

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so

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c-ak = 0

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a+ck = b

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ah wait

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ok so

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the question asks us to show something without the k term

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so we want to cancel it out using these equations

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c = ak

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k = c/a

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k=c/a?

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yup

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ah

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and then solve for k using 2nd equation

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k = (b-a)/c

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so

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c/a = (b-a)/c

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c^2= a(b-a)

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oh damn

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wait

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i think i accidentally swapped b-a

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oops

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either way

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thats how you do it

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im kinda confused on how you swapped p with a and q with c

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p = a
q-pk = 0
p+qk = b
q = c

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from the equation we can see that p=a and q=c

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mb

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sometimes i dont read

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thank you ver ymuch

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thanks @charred rover

junior treeBOT
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@nocturne nest has given 1 rep to @half plank

charred rover
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yw <3

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imma take down your solution