How do you find the polynomial function with leading coefficient 2 that has the given degree and zeroes: degree 3: zeroes -2,1,4?

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How do you find the polynomial function with leading coefficient 2 that has the given degree and zeroes: degree 3: zeroes -2,1,4?

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Answer 1 · 2016-07-20T04:28:18

$2 x^{3} - 6 x^{2} - 12 x + 16$ $2x^3-6x^2-12x+16$ .

Explanation:

The degree of reqd. polynomial, say $p (x)$ $p(x)$ is $3$ $3$ , and hence by the Fundamental Principle of Algebra, it must have $3$ $3$ zeroes. These are given to be $- 2, 1 and 4$ $-2,1 and 4$ .

As $- 2$ $-2$ is a zero of $p (x), x - (- 2) = x + 2$ $p(x), x-(-2)=x+2$ must be a factor of $p (x)$ $p(x)$ .

Similarly, other zeroes give us factors $(x - 1) and (x - 4)$ $(x-1) and (x-4)$

Degree of $p (x)$ $p(x)$ is $3$ $3$ , so, $p (x)$ $p(x)$ can not have any other factor except those described above. Of course, $p (x)$ $p(x)$ can have a numerical factor, like $k \neq 0$ $k!=0$ .

In view of above, we can suppose that,

$p (x) = k (x + 2) (x - 1) (x - 4)$ $p(x)=k(x+2)(x-1)(x-4)$ . Expanding the $R . H . S .$ $R.H.S.$ , we have,

$p (x) = k (x^{3} - 3 x^{2} - 6 x + 8)$ $p(x)=k(x^3-3x^2-6x+8)$

But this will give us the leading co-eff $= k$ $=k$ , which is given to be $2$ $2$ , so, $k = 2$ $k=2$ .

Hence the poly. $p (x) = 2 (x^{3} - 3 x^{2} - 6 x + 8) = 2 x^{3} - 6 x^{2} - 12 x + 16$ $p(x)=2(x^3-3x^2-6x+8)=2x^3-6x^2-12x+16$ .

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How do you find the polynomial function with leading coefficient 2 that has the given degree and zeroes: degree 3: zeroes -2,1,4?

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