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Pre-Calculus Homework help.?

I need help on two problems that are similar. I have the rest of the homework done but these two are giving me problems! if you could solve them or help me step by step that'd be great.

Given a complex polynomial function whose coefficients are real numbers,

a) find the remaining zeros of f, then

b) find a polynomial, in expanded form, w/real coefficients that has zeros

4) degree 3; zeros: 4+i, 6

5) Degree 4; zeros: 1, 2, 2+i

3 Answers

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  • Faz
    Lv 7
    1 decade ago
    Favorite Answer

    Things you need to know

    1) If 'a' is a zero, then (x-a) is a factor

    2) Complex roots always occur in conjugate pairs, so if (a+i) is a factor, then so is (a-i), where i is the imaginary number

    3) i² = -1

    4) Expand

    (x-(4+i)) (x-(4-i)) (x-6)

    = x³ -14x² +65x - 102

    5) Expand

    (x-1)(x-2) (x-(2+i)) (x-(2-i))

    = x⁴ - 7x³ + 19x² - 23x + 10

  • 1 decade ago

    Given a complex polynomial function whose coefficients are real numbers,

    a) find the remaining zeros of f, then

    b) find a polynomial, in expanded form, w/real coefficients that has zeros

    4) degree 3; zeros: 4+i, 6

    Since 4 + i is a zero, 4 – i is also a zero

    Quadratic equation with 4 + i and 4 – i as zeroes:

    x² – ((4 + i) + (4 – i))x + (4 + i)(4 – i)

    x² – 8x + 17

    Other factor is x – 6

    f(x) = (x – 6)(x² – 8x + 17)

    ..... = x(x² – 8x + 17) – 6(x² – 8x + 17)

    ..... = x³ – 8x² + 17x – 6x² + 48x – 102

    ..... = x³ – 14x² + 65x – 102

    5) Degree 4; zeros: 1, 2, 2+i

    Since 2 + i is a zero, 2 – i is also a zero

    Quadratic equation with 2 + i and 2 – i as zeroes:

    x² – ((2 + i) + (2 – i))x + (2 + i)(2 – i)

    x² – 4x + 5

    Quadratic equation with 1 and 2 as zeroes

    x² – (1 + 2)x + (1)(2)

    x² – 3x + 2

    f(x) = (x² – 4x + 5)(x² – 3x + 2)

    ..... = x²(x² – 3x + 2) – 4x(x² – 3x + 2) + 5(x² – 3x + 2)

    ..... = x⁴ – 3x³ + 2x² – 4x³ + 12x² – 8x + 5x² – 15x + 10

    ..... = x⁴ – 3x³ – 4x³ + 2x² + 12x² + 5x² – 8x – 15x + 10

    ..... = x⁴ – 7x³ + 19x² – 23x + 10

  • ?
    Lv 4
    5 years ago

    V = (s)(s)(h) = one hundred SA = (s)(s) + 4(s)(h) = one hundred twenty from the 1st equation you get h = one hundred/(s*s) plug that into the 2d equation (s)(s) + 4(one hundred)/s = one hundred twenty multiply each and every thing with the aid of s and you get s^3 + 4 hundred = 120s you will have a graphing calculator, with the aid of which you will discover the dimensions of the area after which the top of the field to discover the optimum quantity V = (s)(s)(h) = max SA = (s)(s) + 4(s)(h) = one hundred twenty from the 2d equation you get h = (one hundred twenty - s^2) / (4s) plug it into the 1st equation V = (s^2)(one hundred twenty - s^2) / (4s) simplify with the aid of doing away with an s V = 30s- (s^3)/4 so plug that into your graphing calculator and get your optimum

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