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Leading Coefficient 2, and having zeros 3 and -4.

2006-10-15 14:55:01 · 2 answers · asked by cdb2006 2 in Education & Reference Homework Help

2 answers

y=2(x-3)(x+4)
y=2x^2+2x-24

j

2006-10-16 09:43:57 · answer #1 · answered by odu83 7 · 0 0

i think the two have multiplicity 2? precise the surely factors are undemanding, particularly that (x-5)(x-5) = x^2 -10x + 25 is a polynomial with roots 5, multiplicity 2. Now you won't be in a position to have 3+2i ensue with multiplicity 2, because it is conjugate could prefer to additionally be a root. yet you will locate that the polynomial that has 3+/- 2i as roots is x^2-6x+13 subsequently your polynomial is (x^2-10x +25)(x^2-6x+13) as quickly as you're thinking the thank you to get the 2d polynomial, it is undemanding. you realize that the muse could prefer to be x = 3+2i, so do right here math to get a polynomial (x-3) = 2i (x-3)² = -4 x^2 - 6x +9 = -4 x² - 6x + 13= 0

2016-12-26 20:15:11 · answer #2 · answered by Anonymous · 0 0

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