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Find the zeros of function r(x)=(x-7)^2(x^2+7)

User Greim
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Explanation:

a product of factors is only 0, if at least one of the factors is 0.

as 0 × something = 0 (always).

so, finding the zeros of

r(x) = (x - 7)² × (x² + 7)

this polynomial is of degree 4 (ultimately this will multiply x² by x² = x⁴ as highest exponent for x) and has therefore 4 zeros.

the first term is

(x - 7)²

the zero here counts twice (due to the square). in reality the curve will touch the x-axis there and not intercept it.

when will it be 0 ?

at x = 7, as 7 - 7 = 0.

so,

x = 7 are the first 2 zeroes.

now for the other term

(x² + 7)

when will that be 0 ?

when x² = -7

and that means x1 = +sqrt(7)×i

x2 = - sqrt(7)×i

these 2 zeroes are complex (or imaginary) numbers, and you cannot find them on the curve on a grid for real numbers.

but they exist and are zeroes.

User Peteches
by
7.3k points

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