Find the polynomial of minimum degree, with real coefficients, zeros at x=4+4i and x=2, and y-intercept at 64

Question

Find the polynomial of minimum degree, with real coefficients, zeros at x=4+4i and x=2, and y-intercept at 64

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Trung Dũng 3 years 2021-08-21T20:54:05+00:00 1 Answers 11 views 0

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    2021-08-21T20:55:44+00:00

    Answer:

    \displaystyle -x^3+10x^2-48x+64

    Step-by-step explanation:

    We want to find the minimum-degree polynomial with real coefficients and zeros at:

    x= 4+4i\text{ and }  x = 2

    As well as a y-intercept of 64.

    By the Complex Root Theorem, if a + bi is a root, then abi is also a root.

    So, a third root will be 4 – 4i.

    The factored form of a polynomial is given by:

    P(x)=a(x-p)(x-q)...

    Where a is the leading coefficient and p and q are the zeros. More factors can be added if necessary.

    Substitute:

    P(x)=a(x-(2))(x-(4+4i))(x-(4-4i))

    Since we want the minimum degree, we won’t need to add any exponents.

    Expand the second and third factors:

    \displaystyle \begin{aligned} (x-(4+4i))(x-(4-4i))&=(x-4-4i)(x-4+4i) \\ &= x(x-4-4i)-4(x-4-4i)+4i(x-4-4i)\\ &=x^2-4x-4ix-4x+16+16i+4ix-16i-16i^2\\ &= x^2-8x+32\end{aligned}

    Hence:

    P(x)=a(x-2)(x^2-8x+32)

    Lastly, we need to determine a. Since the y-intercept is y = 64, this means that when x = 0, y = 64. Thus:

    64=a(0-2)(0^2-8(0)+32)

    Solve for a:

    -64a=64\Rightarrow a=-1

    Our factored polynomial is:

    P(x)=-(x-2)(x^2-8x+32)

    Finally, expand:

    \displaystyle \begin{aligned} P(x) &=-(x^2(x-2)-8x(x-2)+32(x-2)) \\&=-(x^3-2x^2-8x^2+16x+32x-64)\\&=-(x^3-10x^2+48x-64)\\&= -x^3+10x^2-48x+64\end{aligned}

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