Polynomials Questions in English

(D) To find the digit in the unit's place,first convert the fraction into a decimal number.
$\frac{151}{100} = 1.51$
In the decimal number $1.51$,the digit to the left of the decimal point is the unit's place.
Therefore,the digit in the unit's place is $1$.

(B) Given $U_{n} = \frac{1}{n} - \frac{1}{n+1}$.
This is a telescoping series.
The sum $S_{5} = U_{1} + U_{2} + U_{3} + U_{4} + U_{5}$ can be written as:
$S_{5} = (\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{5}) + (\frac{1}{5} - \frac{1}{6})$
Observing the terms,the intermediate fractions cancel out:
$S_{5} = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + \frac{1}{5} - \frac{1}{6}$
$S_{5} = 1 - \frac{1}{6}$
$S_{5} = \frac{6-1}{6} = \frac{5}{6}$

(B) Given $x=11.$
We can rewrite the expression $x^{5}-12 x^{4}+12 x^{3}-12 x^{2}+12 x-1$ by substituting $12$ with $(x+1)$ since $x=11$ implies $12=x+1.$
$=x^{5}-(x+1)x^{4}+(x+1)x^{3}-(x+1)x^{2}+(x+1)x-1$
$=x^{5}-x^{5}-x^{4}+x^{4}+x^{3}-x^{3}-x^{2}+x^{2}+x-1$
$=x-1$
Substituting $x=11$ into the simplified expression:
$=11-1 = 10.$

(C) Given that $p=99.$
We need to find the value of $p(p^{2}+3p+3).$
Expanding the expression,we get $p^{3}+3p^{2}+3p.$
To simplify this,we add and subtract $1$ to complete the cube formula $(a+b)^{3} = a^{3}+3a^{2}b+3ab^{2}+b^{3}.$
So,$p^{3}+3p^{2}+3p+1-1 = (p+1)^{3}-1.$
Substituting $p=99,$ we get $(99+1)^{3}-1 = (100)^{3}-1.$
$(100)^{3} = 1000000.$
Therefore,$1000000-1 = 999999.$