Find a quadratic function with zeros -7i and 7i

Answer 1

(x-7i)(x+7i) = x^2+49

Answer 2

Use the strategy to solve the higher degree polynomial equation and find the roots of the equation.

6x3 + 14x2 - 3x - 7 = 0

Answer 3

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?

Answer 4

If f(x) is a polynomial function of degree 3 and has zeros of
1 and -4i, then find the third zero.

Answer 5

complex zeros come in conjugate pairs. Just as in the first problem above. So, what do you think?

Answer 6

@ 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.

Answer 7

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

equation:
x^3 - x^2 + 16x- 16 = 0

Answer 8

Now i get it. Thanks a lot

Answer 9

find the zeros and multiplicity of the equation, Range and interval q is positive.
q(x)=-x^2(x+1)^2(x-2)