$\bf BC=2BA\\ BE = 2BD \end{cases} \cfrac{BC}{BE}=\cfrac{~~\begin{matrix} 2 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~BA}{~~\begin{matrix} 2 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~BD}\implies \cfrac{BC}{BE}=\cfrac{BA}{BD}\impliedby \begin{array}{llll} \textit{equal ratio of sides}\\ \textit{similar triangles} \end{array}$

8 0

3 years ago

Given that −4i is a zero, factor the following polynomial function completely. Use the Conjugate Roots Theorem, if applicable. f

GalinKa [24]

Answer:

$\large \boxed{\sf \bf \ \ f(x)=(x-4i)(x+4i)(x+3)(x-5) \ \ }$

Step-by-step explanation:

Hello, the Conjugate Roots Theorem states that if a complex number is a zero of real polynomial its conjugate is a zero too. It means that (x-4i)(x+4i) are factors of f(x).

$\text{Meaning that } (x-4i)(x+4i) =x^2-(4i)^2=x^2+16 \text{ is a factor of f(x).}$

The coefficient of the leading term is 1 and the constant term is -240 = 16 * (-15), so we a re looking for a real number such that.

$f(x)=x^4-2x^3+x^2-32x-240\\\\ =(x^2+16)(x^2+ax-15)\\\\ =x^4+ax^3-15x^2+16x^2+16ax-240$

We identify the coefficients for the like terms, it comes

a = -2 and 16a = -32 (which is equivalent). So, we can write in $\mathbb{R}$ .

$\\f(x)=(x^2+16)(x^2-2x-15)$

The sum of the zeroes is 2=5-3 and their product is -15=-3*5, so we can factorise by (x-5)(x+3), which gives.

$f(x)=(x^2+16)(x^2-2x-15)\\\\=(x^2+16)(x^2+3x-5x-15)\\\\=(x^2+16)(x(x+3)-5(x+3))\\\\=\boxed{(x^2+16)(x+3)(x-5)}$

And we can write in $\mathbb{C}$

$f(x)=\boxed{(x-4i)(x+4i)(x+3)(x-5)}$

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

7 0

3 years ago