(x-7i)(x+7i) = x^2+49
Use the strategy to solve the higher degree polynomial equation and find the roots of the equation.
6x3 + 14x2 - 3x - 7 = 0
6 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?
If f(x) is a polynomial function of degree 3 and has zeros of
1 and -4i, then find the third zero.
complex zeros come in conjugate pairs. Just as in the first problem above. So, what do you think?
@ 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.
Vikky, 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
x^3 - x^2 + 16x- 16 = 0
Now i get it. Thanks a lot
find the zeros and multiplicity of the equation, Range and interval q is positive.
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