If \[{{x}^{3}}-5x+7\] is divisible by \[(x+2)\], then the remainder is

1.	If \[{{x}^{3}}-5x+7\] is divisible by \[(x+2)\], then the remainder is
A.	-21
B.	-20
C.	-17
D.	-25
Answer» B. -20

Discussion

No Comment Found

Related MCQs

The product of the polynomials, \[(3{{u}^{2}}v-5u{{v}^{2}})\] and \[\left( \frac{1}{5}{{u}^{2}}+\frac{1}{5}{{v}^{2}} \right)\] is
If \[a+b+c=0\], the value of \[\frac{{{a}^{2}}}{bc}+\frac{{{b}^{2}}}{ca}+\frac{{{c}^{2}}}{ab}\]is
The expansion of log \[{{\log }_{e}}\frac{1}{(1-x)}\]is
Read the following statements carefully and choose the correct answer. (i) I had Rs. 200 with me. I gave Rs. x to Anwar, Rs. \[\frac{x}{2}\] to Vidhu and I am left with Rs. \[\frac{x}{2}\]. The algebraic form of the given statement is written as: \[x+\frac{x}{2}+\frac{x}{2}=200\] (ii) The amount I gave to Vidhu is Rs 50 and to Anwar is Rs 100.
For which of the following function is : \[\frac{f-f}{a-b}\] Constant for all number a and b where \[a\ne b\]?
DIRECTIONS: The questions in this segment consists of two statements, one labelled as ?Assertion A? and the other labelled as ?Reason R?. You are to examine these two statements carefully and decide if the Assertion A and Reason R are individually true and if so, whether the reason is a correct explanation of the assertion. Select your answers to these items using codes given below. Assertion (A): Value of \[{{\left( \frac{4}{9} \right)}^{\frac{3}{2}}}\times {{\left( \frac{4}{9} \right)}^{\frac{1}{2}}}\]is \[{{\left( \frac{4}{9} \right)}^{2}}.\] Reason (R): For any two rational numbers a and b and for any integers m and n, we have \[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}\]
If \[x-\frac{1}{x-2}=2-1\frac{1}{x-2}\] , then x is equal to
What is the value of "n" that makes the equation below true? \[\frac{n}{3}=12\].
The equation \[\frac{(x+2)(x-5)}{(x-3)(x+6)}=\frac{x-2}{x+4}\] has
If \[x+\frac{1}{x}=5\] (where\[x\ne 0\]) then find \[\frac{2x}{3{{x}^{2}}-5x+3}=?\]

If \[{{x}^{3}}-5x+7\] is divisible by \[(x+2)\], then the remainder is

Discussion

No Comment Found

Related MCQs

Reply to Comment