I'm stuck on this problem... anybody know what the answer is?
1 answer:
Angles A and D would have to be congruent in order to prove SAS Similarity Theorem.
“The SAS Similarity Theorem states that if two sides in one triangle are proportional to two sides in another triangle and the included angle in both are congruent, then the two triangles are similar.” (Source: cK-12)
This graphic may help you understand my choice of answer. It shows SAS Similarity Theorem in an alternate example.
“The SAS Similarity Theorem states that if two sides in one triangle are proportional to two sides in another triangle and the included angle in both are congruent, then the two triangles are similar.” (Source: cK-12)
This graphic may help you understand my choice of answer. It shows SAS Similarity Theorem in an alternate example.
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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).