How do you write a polynomial function of least degree and leading coefficient 1 when the zeros are 0, 2, 2, 4?

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Answer 1 · 2016-09-12T04:28:30

$x^{4} - 8 x^{3} + 20 x^{2} - 16 x = 0$ $x^4-8x^3+20x^2-16x=0$

We'll start with the zeros:

$x = 0, x = 2, x = 2, x = 4$ $x=0, x=2, x=2, x=4$

These are called "zeros" because we're looking for where the graph crosses the X-axis - or in other words, where "y=0". So we'll have:

$x = 0, x - 2 = 0, x - 2 = 0, x - 4 = 0$ $x=0, x-2=0, x-2=0, x-4=0$

We do this because in a function, if one term within an equation that is all multiplication equals 0, the entire function equals 0. So we get:

$x (x - 2) (x - 2) (x - 4) = 0$ $x(x-2)(x-2)(x-4)=0$

And now it's just expanding all these terms into one polynomial equation:

$x (x^{2} - 4 x + 4) (x - 4) = 0$ $x(x^2-4x+4)(x-4)=0$

$x (x^{3} - 8 x^{2} + 20 x - 16) = 0$ $x(x^3-8x^2+20x-16)=0$

$x^{4} - 8 x^{3} + 20 x^{2} - 16 x = 0$ $x^4-8x^3+20x^2-16x=0$