- #1

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the solution:

identify the easy root of x=1,

find the remaining roots from (x-1)(x^2+4x) using quadratic formula.

The only thing i dont understand here is how to factorize to (x-1)(x^2+4x)... namely the (x^2+4x) part.

- Thread starter stat643
- Start date

- #1

- 16

- 0

the solution:

identify the easy root of x=1,

find the remaining roots from (x-1)(x^2+4x) using quadratic formula.

The only thing i dont understand here is how to factorize to (x-1)(x^2+4x)... namely the (x^2+4x) part.

- #2

- 1,752

- 1

factor out a common term in x^2+4x

and you're pretty much done

and you're pretty much done

- #3

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but how did i get to x^2+4x in the first place?... the original equation was x^3 + 3x^2 - 10x + 6.. i merely copied the solution... so i find the easy root of 1.. then what?

Last edited:

- #4

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use synthetic division

- #5

matt grime

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- #6

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i just looked up synthetic devision on wikipedia and tried it but it didnt work

- #7

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should i take out the common term x first?

- #8

arildno

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Then you should practice synthetic division once more!i just looked up synthetic devision on wikipedia and tried it but it didnt work

Further, ponder over matt grime's words:

WHY should you be suspicious of that particular factorization?

Hint:

How could you ascertain whether the factorization is correct or false?

- #9

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oh sorry i copied it wrong, it should be (x-1)(x^2+4x-6).. now expanding that get: x^3 + 4x^2 -6x -x^2 -4x + 6 = x^3 + 3x^2 - 10x + 6.. so yeh its right now.. though i still cant get the synthetic devision right (its new to me)

i tried to learn it now from http://en.wikipedia.org/wiki/Synthetic_division

though i keep getting 1,2,-12,18

can someone help show how i would use synthetic devision for the original polynomial ?

i tried to learn it now from http://en.wikipedia.org/wiki/Synthetic_division

though i keep getting 1,2,-12,18

can someone help show how i would use synthetic devision for the original polynomial ?

Last edited:

- #10

arildno

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[tex](x-1)(ax^{2}+bx+c)=x^{3}+3x^{2}-10x+6[/tex] holds for all x.

I.e, we must determine a,b and c!!

Multiplying out the left-hand side, and organizing in powers of x, the lefthandside can be rewritten as:

[tex]ax^{3}+(b-a)x^{2}+(c-b)x-c= x^{3}+3x^{2}-10x+6[/tex]

NOw, the coefficients of each power must be equal on the right and left sides, yielding the system of equations:

a=1

b-a=3

c-b=-10

-c=6

This yields:

a=1

b=4

c=-6

- #11

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thanks arildno, very helpful

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