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Which of the following quartic functions has x=-1 and x = -3 as its only two real zeros?

y=x^4 - 4x^3 - 4x^2 - 4x - 3

y = -x^4 + 4x^3 + 4x^2 + 4x + 3

y x^4 + 4x^3 + 3x^2 + 4x

x^4 + 4x^3 + 4x^2 + 4x + 3

User Jacob Wu
by
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1 Answer

1 vote

Answer:


y=x^4 + 4x^3 + 4x^2 + 4x + 3

Explanation:

Consider function
y=x^4 + 4x^3 + 4x^2 + 4x + 3.

Factor it


y=x^4 + 4x^3 + 4x^2 + 4x + 3\\ \\=x^4+x^3+3x^3+3x^2+x^2+x+3x+3\\ \\=x^3(x+1)+3x^2(x+1)+x(x+1)+3(x+1)\\ \\=(x+1)(x^3+3x^2+x+3)\\ \\=(x+1)(x^2(x+3)+(x+3))\\ \\=(x+1)(x+3)(x^2+1)

This function has two real zeros
x=-1 and
x=-3 and two comples zeros (because
x^2+1 cannot be factored futher).

Hence, the function
y=x^4 + 4x^3 + 4x^2 + 4x + 3 has
x=-1 and
x=-3 as its only two real zeros.

User Chanaka
by
8.1k points
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