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P1 Question.....

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Reply 20
Avatar for Bhaal85
Bhaal85
9
Bhaal85
If the equation is x^3-19x-30, therefore:

(X+2)(X^2-2X-15) which can be further factorised to give:

(X+2)(X+3)(X-5)


That is the answer, just done it. If you need someone to elaborate feel free to ask bono or Elpaw.
Reply 21
Avatar for elpaw
elpaw
15
by algebraic juggling:


(x³-19x-30)/(x+2)

= (x²(x+2))/(x+2) -2x²/(x+2) -(19x+30)/(x+2)

= x² - (2x(x+2))/(x+2) + 4x/(x+2) -(19x + 30)/(x+2)

= x² - 2x - 15(x+2)/(x+2) + (30 - 30)/(x+2)

= x² -2x -15

= (x-5)(x+3)
Reply 22
Avatar for elpaw
elpaw
15
eastldn
it was on the P1 JAN 04 PAPER, 1ST QUESTION..... (x+2) is a factor
f(x) = X^3 - 19x - 30
f(-2) = 0
im sure it can be factorised, bt how?

you said f(x) = x^3 - 19x^2 - 30 originally. that can't be factorised by (x+2)

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