How to factor this polynominal?

[hkus10]

 

1) (x-10)[(x+4)(x+1) - 24] - 3[(-11x - 11) + 24] + 8[-21 + 3x]

 

what I get is (x-10)(x^2+5x-20) + 57x-207

The reason that I do not combine them is because I think it is much more difficult to deal with x^3?

What should I do here?

[imatfaal]

Whilst some equations will allow you pluck a factor straight out from an equation with two or more sets of brackets added together, most of the time you need to carefully multiply out completely, collect similar terms, and start from there. the needless complexity of the above makes me think that this long winded route is exactly what the setter wanted to test. make sure you follow rules for brackets and multiplication closely and keep firm track of minus signs.

 

Edit

 

after a bit of crunching - this is not a bad one! Follow method above - factors/roots should be easily guessable once you have simple cubic. Remember that the term with no powers of x must be the multiple of all of the three numbers in the brackets ie

 

y=(x+a)(x+b)(x+c) = x^3 + (a+b+c)x^2 + (ab+ac+bc)x + abc

 

In this case there are only a very small number of ways of making abc

Edited May 5, 2011 by imatfaal

[imatfaal]

Hi Sofia

 

That's not the expansion that I got - and just as importantly the question asks for correct factorisation not expansion. It doesn't look like HKUS is coming back so I don't mind answering his homework now

 

(x-10)[(x+4)(x+1) - 24] - 3[(-11x - 11) + 24] + 8[-21 + 3x] I have put in extra spaces to exaggerate the sections

(x-10)[(x^2+5x+4)-24] +33x+33-72 + 24x -168 (multiply out innermost bracket (x+4)(x+1) - and simplify other terms

(x-10)[(x^2+5x-20] +33x+33-72 + 24x -168 (add in constant)

x^3+5x^2-20x-10x^2-50x+200 +33x+33-72 + 24x -168 (multiply bracket by (x-10)

x^3 + 5x^2 - 10x^2 -20x-50x+33x+24x +200-72-168+33 (gather like terms)

X^3 -5x^2 -13x -7

To factorise you know you need three constants that multiply together to give 7 (ie its gotta be 7 and two ones - but you need to check minus signs) a little work gives

(x+1)(x+1)(x-7)

Hope that helps