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alexgriva [62]
2 years ago
13

What is the polynomial function of lowest degree with lead coefficient 1 and roots i, –2, and 2?

Mathematics
1 answer:
lesya [120]2 years ago
3 0

Answer:

x^4 - 3x^2 - 4.

Step-by-step explanation:

Imaginary roots  occur as conjugate pairs  so a 4th roots is -i.

So in factor form we have:

f(x) = (x - 2)(x + 2)(x - i)x + i)

     = (x^2 + 1)(x - 2)(x + 2)

     = (x^2 + 1)(x^2 - 4)

      = x^4 - 3x^2 - 4.

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1 year ago
For what real values of $c$ is $x^2 - 8x + c$ the square of a binomial?
nikklg [1K]

Value of c is 16 for which equation  f(x) = x^2-8x+c  is a square of binomial !

<u>Step-by-step explanation:</u>

Here we meed to find the value of c for which equation f(x) = x^2-8x+c or ,  f(x) = x^2-8x+c is a square of a binomial . Let's find out:

We know that

⇒ (x-a)^2 = x^2-2(a)(x)+a^2   ..........(1)

Let's simplify given equation in question

⇒ f(x) = x^2-8x+c

⇒ f(x) = x^2-2(4)x+c

Comparing this equation with (1) we get :

a=4 , c=a^2

⇒ c=a^2

⇒ c=4^2

⇒ c=16

Therefore , Value of c is 16 for which equation  f(x) = x^2-8x+c  is a square of binomial !

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