Use the given zero to find all the zeroes of the function

[kryms3n]

In these exercises, use the given zero to find all the zeroes of the function.

56) g(x) = 4x^3 + 23x^2 + 34x - 10, with one zero being -3 + i

57) h(x) = 3x^3 - 4x^2 + 14x + 20, with on zero being 1 - radical(3) i

I need some help with these. Thank you!

 

[tkhunny]

You should have learned about "conjugates". Those Integer coefficients are a big help.

Given -3+1, you are supposed to come up with -3-i.

Similarly, given 1-radical (3) i, you are supposed to come up with 1+radical (3) i.

Now what?

 

[stapel]

Follow the advice in the previous reply. For (56), plug the given function and the given zero into synthetic division. Into what results, plug the conjugate. Once you have divided out these two zeroes, you'll be left with a linear factor which you can easily solve. And (57) works the same way.

Eliz.

 

[soroban]

Re: Use the given zero to find all the zeroes of the functio

Hello, kryms3n!

Did you understand all the excellent advice you got?

In these exercises, use the given zero to find all the zeroes of the function.

\(\displaystyle 56)\;g(x)\:=\:4x^3\,+\,23x^2\,+\,34x\,-\,10\), with one zero being -\(\displaystyle 3\,+\,i\)

Complex roots occur in conjugate pairs.
. . If -\(\displaystyle 3\,+\,i\) is a root, then -\(\displaystyle 3\,-\,i\) is also a root.

Then: \(\displaystyle \,x\,-\,(-3\,+\,i)\:=\:x\,+\,3\,-\,i\) is a factor
. and: \(\displaystyle \,x\,-\,(-3\,-\,i)\:=\:x\,+\,3\,+\,i\) is a factor.

Hence: \(\displaystyle \,(x\,+\,3\,-\,i)(x\,+\,3\,+\,i)\:=\:x^2\,+\,6x\,+\,10\) is a factor.

Use long division: \(\displaystyle \,(4x^3\,+\,23x^2\,+\,34x\,-\,10)\,\div\,(x^2\,+\,6x\,+\,10) \;= \;4x\,-\,1\)


And we have: \(\displaystyle \:g(x)\;=\;\underbrace{\left(x\,-\,[-3\,+\,i]\right)}_{x\,=\,-3+i}\underbrace{\left(x\,-\,[-3\,-\,i]\right)}_{x\,=\,-3-i}\underbrace{\left(4x\,-\,1\right)}_{x\,=\,\frac{1}{4}}\)

\(\displaystyle 57)\;h(x)\:=\:3x^3\,-\,4x^2\,+\,14x\,+\,20\), with on zero being \(\displaystyle \,1\,-\,\sqrt{3}\)

Irrational roots also occur in conjugate pairs.
. . If \(\displaystyle 1\,-\,\sqrt{3}\) is a root, then \(\displaystyle 1\,+\,\sqrt{3}\) is also a root.

Then: \(\displaystyle x\,-\,(1\,-\,\sqrt{3})\:=\:x\,-\,1\,+\,\sqrt{3}\) is a factor
. and: \(\displaystyle x\,-\,(1\,+\,\sqrt{3})\:=\:x\,-\,1\,-\,sqrt{3}\) is a factor.

Hence: \(\displaystyle \,(x\,-\,1\,+\,\sqrt{3})(x\,-\,1\,-\,\sqrt{3})\:=\:x^2\,-\,2x\,-\,2\) is a factor.

Can you finish it now?