Give an example and explain why a polynomial can have fewer x-intercepts than its number of roots.
1 answer:
Answer:
A polynomial can have fewer x-intercepts than its number of roots when a pair of complex conjugate roots exist.
Example:
Consider the 4-th degree polynomial
f(x) = x⁴ - x³ - x² - x - 2
According to the Remainder theorem
f(-1) = 1 + 1 - 1 + 1 - 2 = 0
Therefore (x + 1) is a factor.
f(2) = 16 - 8 - 4 - 2 - 2 = 0
Therefore (x - 2) is a factor.
(x+1)*(x-2) = x² - 2x + x - 2 = x² - x - 2
To find the remaining factor, perform long division.
x² + 1
-------------------------------
x²-x-2 | x⁴ - x³ - x² - x - 2
x⁴ - x³ - 2x²
-----------------------------
x² - x - 2
x² - x - 2
Therefore
f(x) = (x+1)(x-2)(x²+1)
Notice that (x² + 1) has no real factors.
However,
x² + 1 = (x + i)(x - i),
so it has a pair of conjugate zeros +i and -i.
A graph of the function confirms that there are only two real zeros (shown in red color).
A polynomial can have fewer x-intercepts than its number of roots when a pair of complex conjugate roots exist.
Example:
Consider the 4-th degree polynomial
f(x) = x⁴ - x³ - x² - x - 2
According to the Remainder theorem
f(-1) = 1 + 1 - 1 + 1 - 2 = 0
Therefore (x + 1) is a factor.
f(2) = 16 - 8 - 4 - 2 - 2 = 0
Therefore (x - 2) is a factor.
(x+1)*(x-2) = x² - 2x + x - 2 = x² - x - 2
To find the remaining factor, perform long division.
x² + 1
-------------------------------
x²-x-2 | x⁴ - x³ - x² - x - 2
x⁴ - x³ - 2x²
-----------------------------
x² - x - 2
x² - x - 2
Therefore
f(x) = (x+1)(x-2)(x²+1)
Notice that (x² + 1) has no real factors.
However,
x² + 1 = (x + i)(x - i),
so it has a pair of conjugate zeros +i and -i.
A graph of the function confirms that there are only two real zeros (shown in red color).
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