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August 29, 2016
Posted by **Vikky** on Wednesday, July 2, 2014 at 7:39pm.

- algebra -
**Steve**, Wednesday, July 2, 2014 at 7:41pm(x-7i)(x+7i) = x^2+49

- algebra -
**Vikky**, Wednesday, July 2, 2014 at 8:01pmUse the strategy to solve the higher degree polynomial equation and find the roots of the equation.

6x3 + 14x2 - 3x - 7 = 0 - algebra -
**Damon**, Wednesday, July 2, 2014 at 8:07pm6 x^3 - 3 x + 14 x^2 - 7 = 0

3 x (2 x^2 - 1) + 7 (2 x^2-1) = 0

(3x+7)( 2x^2-1) = 0

does that help? - algebra -
**Vikky**, Wednesday, July 2, 2014 at 8:29pmIf f(x) is a polynomial function of degree 3 and has zeros of

1 and -4i, then find the third zero. - algebra -
**Steve**, Wednesday, July 2, 2014 at 8:36pmcomplex zeros come in conjugate pairs. Just as in the first problem above. So, what do you think?

- algebra -
**Vikky**, Wednesday, July 2, 2014 at 9:11pm@ Steve. i still don't get it. should i take the "i" out of the 4? because if i do something like this, (x-)(x+4i) i will have 4ix, of which its confusing.

- algebra -
**Reiny**, Wednesday, July 2, 2014 at 10:09pmVikky, what Steve is telling you that complex roots, as well as irrational roots, always come in conjugate pairs.

e.g. if one root is 5 - 7i, then there has to be a matching 5 + 7i

e.g. if one root is (4 + 3√5)/2 , then there has to be a (4 - 3√5)/2

So, if you know one root is -4i , then there has to be a +4i

Furthermore, since 1 is also a root, we have 3 factors,

(x-1)(x + 4i)(x - 4i)

= (x-1)(x^2 - 16i^2)

= (x-1)(x^2 + 16)

= x^3 - x^2 + 16x - 16

equation:

x^3 - x^2 + 16x- 16 = 0 - algebra -
**Vikky**, Wednesday, July 2, 2014 at 10:54pmNow i get it. Thanks a lot

- algebra -
**solomon**, Friday, July 4, 2014 at 1:08pmfind the zeros and multiplicity of the equation, Range and interval q is positive.

q(x)=-x^2(x+1)^2(x-2)