NO LINKS!! Find an nth-degree polynomial function with real coefficient satisfying the given conditions.

Mathematics

2 answers:

Sati [7]2 years ago

6 0

n=3; We need a third degree polynomials with the following given zero's: 2 and 5i are zeros; f(-1)=156.

Since these are solutions

x = 2 ; x = 5i. Since imaginaries travel in pairs, the other answer is x= -5i.

We have (x-2)(x-5i)(x+5i) = 0

Now,

f(-1) = (-1-2)(-1-5i)(-1+5i) = 156.

f(-1) = (-3)(26) = -78.

But -78 x -2 = 156, so our polynomial becomes

Y= -2x (x - 2 ) x (x to the power of 2 + 25) = 0

marissa [1.9K]2 years ago

4 0

Answer:

$f(x)=2x^4-12x^3+50x^2-56x-120$

Step-by-step explanation:

Complex Conjugate Theorem

For a polynomial p(x) with real coefficients, the complex zeros occur in conjugate pairs. So if (a + bi) is a zero, then its conjugate (a - bi) is also a zero.

Given zeros:

-1, 3, and (2 + 4i)

As one of the given zeros is a complex number, (2 - 4i) is also a zero.

Write each zero as part of a factor and multiply them together, adding a leading coefficient $a$ :

$\begin{aligned}f(x) & =a(x+1)(x-3)(x-(2+4i))(x-(2-4i))\\& = a(x+1)(x-3)(x-2-4i)(x-2+4i)\\& = a(x+1)(x-3)(x^2-2x+4ix-2x+4-8i-4ix+8i-16i^2)\\& = a(x+1)(x-3)(x^2-4x+4-16i^2)\end{aligned}$

Remember that $i^2=-1$ , therefore:

$\begin{aligned}f(x)&=a(x+1)(x-3)(x^2-4x+4-16(-1))\\&=a(x+1)(x-3)(x^2-4x+4+16)\\&=a(x+1)(x-3)(x^2-4x+20)\end{aligned}$

To find the value of the leading coefficient (a), use the given $f(1)=-136$ :

$\begin{aligned}f(1) & = -136\\\implies a(1+1)(1-3)(1^2-4(1)+20) & = -136\\a(2)(-2)(17) & = -136\\-68a & = -136\\\implies a & = 2\end{aligned}$

Therefore, the polynomial in factored form is:

$f(x)=2(x+1)(x-3)(x^2-4x+20)$

Finally, expand the brackets:

$\begin{aligned}f(x) & =2(x+1)(x-3)(x^2-4x+20)\\& =2(x^2-2x-3)(x^2-4x+20)\\& =2(x^4-4x^3+20x^2-2x^3+8x^2-40x-3x^2+12x-60)\\& =2(x^4-4x^3-2x^3+20x^2+8x^2-3x^2-40x+12x-60)\\& =2(x^4-6x^3+25x^2-28x-60)\\& =2x^4-12x^3+50x^2-56x-120\end{aligned}$