Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Given the set of zeros 
{-5,1,+-2i}, which is a polynomial of least degree in factored form with a leading coefficient of one?

f(x)=(x-5)(x+1)(x^(2)-4)

f(x)=(x+5)(x-1)(x^(2)+2)

f(x)=(x-5)(x+1)(x+2i)^(2)

f(x)=(x+5)(x-1)(x^(2)+4)

Given the set of zeros {5,1,±2i} \{-5,1, \pm 2 i\} , which is a polynomial of least degree in factored form with a leading coefficient of one?\newlinef(x)=(x5)(x+1)(x24) f(x)=(x-5)(x+1)\left(x^{2}-4\right) \newlinef(x)=(x+5)(x1)(x2+2) f(x)=(x+5)(x-1)\left(x^{2}+2\right) \newlinef(x)=(x5)(x+1)(x+2i)2 f(x)=(x-5)(x+1)(x+2 i)^{2} \newlinef(x)=(x+5)(x1)(x2+4) f(x)=(x+5)(x-1)\left(x^{2}+4\right)

Full solution

Q. Given the set of zeros {5,1,±2i} \{-5,1, \pm 2 i\} , which is a polynomial of least degree in factored form with a leading coefficient of one?\newlinef(x)=(x5)(x+1)(x24) f(x)=(x-5)(x+1)\left(x^{2}-4\right) \newlinef(x)=(x+5)(x1)(x2+2) f(x)=(x+5)(x-1)\left(x^{2}+2\right) \newlinef(x)=(x5)(x+1)(x+2i)2 f(x)=(x-5)(x+1)(x+2 i)^{2} \newlinef(x)=(x+5)(x1)(x2+4) f(x)=(x+5)(x-1)\left(x^{2}+4\right)
  1. Identify Zeros and Factors: Identify the zeros of the polynomial and their corresponding factors.\newlineGiven zeros: 5-5, 11, +2i+2i, 2i-2i\newlineCorresponding factors: (x+5)(x + 5), (x1)(x - 1), (x2i)(x - 2i), (x+2i)(x + 2i)
  2. Complex Zeros Conjugate Pairs: Recognize that complex zeros come in conjugate pairs. Since +2i+2i and 2i-2i are a pair of complex conjugate zeros, their factors multiply to a quadratic polynomial with real coefficients. (x2i)(x+2i)=x2(2i)2=x2+4(x - 2i)(x + 2i) = x^2 - (2i)^2 = x^2 + 4
  3. Combine Factors for Polynomial: Combine all factors to form the polynomial of least degree.\newlinef(x)=(x+5)(x1)(x2+4)f(x) = (x + 5)(x - 1)(x^2 + 4)
  4. Match with Given Options: Check the given options to see which one matches the derived polynomial. f(x)=(x+5)(x1)(x2+4)f(x) = (x + 5)(x - 1)(x^2 + 4) matches the last option.

More problems from Write a quadratic function from its x-intercepts and another point