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Fundamental Theorem of Algebra

The degree of...

Question

The degree of the polynomial function $f (x) = (x - 1)^{2} (x + 2)^{2}$ is $4.$

A

False

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B

True

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Solution

The correct option is B True

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f (x)

be a polynomial function of degree

2

f (x) > 0

x \in R

g (x) = f (x) + f^{^{'}} (x) + f^{''} (x)

x

\int \frac{f (x)}{1 - x^{3}} d x = log ∣ ∣ ∣ \frac{x^{2} + x + 1}{x - 1} ∣ ∣ ∣ \frac{A}{948 \sqrt{3}} t a n^{- 1} \frac{2 x + 1}{\sqrt{3}} + C

f (x)

is a polynomial of second degree in x such that

f (0) = f (1) = 3 f (2) = 3

f (x) = \sqrt{{log}_{10} (\frac{2 [x] + 20}{[x]^{2} - 4})}

[.]

denotes the greatest integer function and

S_{1} = {x | f (x) is defined; x \in I^{+}} S_{2} = {N | S_{2} \subset S_{1}; N \in I^{+}}

Then the maximum degree of polynomial having distinct roots

{S_{2}}

f (x)

is a polynomial function of the second degree such that

f (- 3) = 6, f (0) = 6

f (2) = 11

, then the graph of the function

f (x)

cuts the ordinate

x = 1

f (x) = ⎧ ⎨ ⎩ \begin{matrix} 2 x^{2} + 12 x + 16 & , - 4 \leq x \leq - 2 2 - | x | & , - 2 < x \leq 1 4 x - x^{2} - 2 & , 1 < x \leq 3 \end{matrix}

f (x)

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