Let a be the positive integer and b = 4.
Then, by Euclid’s algorithm, a = 4q + r for some integer q ≥ 0 and r = 0, 1, 2, 3 because 0 ≤ r < 4.
So, a = 4q or 4q + 1 or 4q + 2 or 4q + 3.
{tex}(4q)^3\;=\;64q^3\;=\;4(16q^3){/tex}
= 4m, where m is some integer.
{tex}(4q+1)^3\;=\;64q^3+48q^2+12q+1=4(\;16q^3+12q^2+3q)+1{/tex}
= 4m + 1, where m is some integer.
{tex}(4q+2)^3\;=\;64q^3+96q^2+48q+8=4(\;16q^3+24q^2+12q+2){/tex}
= 4m, where m is some integer.
{tex}(4q+3)^3\;=\;64q^3+144q^2+108q+27{/tex}

=4×(16q³+36q²+27q+6)+3
= 4m + 3, where m is some integer.
Hence, The cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3 for some integer m.

If p(x) is a non zero polynomial and p(k²) = 0
Then, k² is a zero of polynomial p(x) . So, degree of p(x) should be more than or equal to one.
Therefore, least degree of the p(x) is one.

Let {tex} \alpha , \beta , \gamma{/tex} be the zeroes of polynomial f(x) = x³ - 5x² - 2x + 24 such that {tex} \alpha\beta{/tex} = 12.
{tex} \alpha + \beta + \gamma = - \left( - \frac { 5 } { 1 } \right) = 5{/tex} ................ (i)
{tex} \alpha \beta + \beta \gamma + \gamma \alpha = \frac { - 2 } { 1 } = - 2{/tex}
and, {tex} \alpha \beta \gamma = - \frac { 24 } { 1 } = - 24{/tex}
Putting, {tex} \alpha \beta = 12{/tex} in {tex} \alpha \beta \gamma = - 24{/tex}, we get
{tex} 12 \gamma = - 24{/tex}
{tex} \Rightarrow \gamma = - \frac { 24 } { 12 } = - 2{/tex}
Putting {tex} \gamma= -2{/tex} in eq.(i), we get
{tex} \alpha + \beta - 2 = 5{/tex}
{tex} \Rightarrow \quad \alpha + \beta = 7{/tex}
Now, {tex} ( \alpha - \beta ) ^ { 2 } = ( \alpha + \beta ) ^ { 2 } - 4 \alpha \beta{/tex}
{tex} \Rightarrow \quad ( \alpha - \beta ) ^ { 2 } = 7 ^ { 2 } - 4 \times 12{/tex}
{tex} \Rightarrow \quad ( \alpha - \beta ) ^ { 2 } = 1{/tex}
{tex} \Rightarrow \quad \alpha - \beta = \pm 1{/tex}
Thus, we have
{tex} \alpha + \beta= 7{/tex}  and {tex} \alpha - \beta = 1 {/tex} or, {tex} \alpha + \beta= 7{/tex} and {tex} \alpha - \beta= -1{/tex}
CASE I: When {tex} \alpha + \beta= 7{/tex} and {tex} \alpha - \beta= 1{/tex}
Solving {tex} \alpha + \beta= 7{/tex} and {tex} \alpha - \beta= 1{/tex} , we get
{tex} \alpha = 4{/tex} and {tex}\beta= 3{/tex} 
CASE II: When {tex} \alpha + \beta= 7{/tex} and {tex} \alpha - \beta= -1{/tex}
Solving {tex} \alpha + \beta= 7{/tex} and {tex} \alpha - \beta= -1{/tex} , we get
{tex} \alpha = 3{/tex} and {tex}\beta= 4{/tex} .
Hence, the zeros of the given polynomial are 3, 4 and - 2 or 4,3 , -2.

Let n be an arbitrary positive integer.
On dividing n by 3, let m be the quotient and r be the remainder.
The positive integer is in the form of 3m or (3m + 1) or (3m + 2)
Then, by Euclid's division lemma, we have
n = 3m + r, where {tex}0 \leq r < 3{/tex}.
{tex}\therefore{/tex} n = 3m or (3m +1) or (3m + 2), for some integer m.
Thus, any positive integer is of the form 3m or (3m + 1) or (3m + 2) for some integer m.

Let the present ages of Aftab and his daughter be x year and y year respectively. Then the algebraic representation is
Given by the following equations:
x - 7 = 7(y - 7)
{tex}\Rightarrow{/tex} x - 7y + 42 = 0 ...(1)
And x + 3 = 3(y + 3)
{tex}\Rightarrow{/tex} x - 3y - 6 = 0 ...(2)
To, represent this equation graphically, well find two solution for each equation, These solution are given below;
For Equation (1) x - 7y + 42 = 0
{tex}\Rightarrow{/tex} 7y = x + 42
{tex}\Rightarrow y = \frac{{x + 42}}{7}{/tex}
Table 1 of solutions

For Equation (2) x - 3y - 6 = 0
{tex}\Rightarrow{/tex} 3y = x - 6
 {tex}\Rightarrow y = \frac{{x - 6}}{3}{/tex}
Table 2 of solutions

  We plot the A(0, 6) and B(7, 7)
Corresponding to the solutions in table 1 on a graph paper to get the line AB representing the equation (1) and the points C(0, -2) and D(6, 0) corresponding to the solutions in table 2 on the same graph paper to get the line CD representing the equation (2), as shown in the figure

We observe in figure that the two lines representing the two equations are intersecting at the point P(42, 12).

Given, 2x² - 7x - 15
= 2x² - 10x + 3x - 15
{tex}= 2x (x - 5) + 3(x - 5)\\= (x - 5)(2x + 3){/tex}

Let usual speed = x km/hr
New speed = (x + 250)km/hr
Total distance = 1500 km
Time taken by usual speed = {tex}\frac{{1500}}{x}{/tex}hr
Time taken by new speed = {tex}\frac{{1500}}{{x + 250}}{/tex}hr
According to question,
{tex}\frac { 1500 } { x } - \frac { 1500 } { x + 250 } = \frac { 1 } { 2 }{/tex}
{tex}\Rightarrow \frac { 1500 x + 1500 \times 250 - 1500 x } { x ^ { 2 } + 250 x } = \frac { 1 } { 2 }{/tex}
{tex}\Rightarrow{/tex} x² + 250x = 750000
{tex}\Rightarrow{/tex} x² + 250x - 750000 = 0
{tex}\Rightarrow{/tex} x² + 1000x - 750x - 750000 = 0
{tex}\Rightarrow{/tex} x(x + 1000) - 750(x + 1000) = 0
{tex}\Rightarrow{/tex} x = 750 or x = -1000
Therefore, usual speed is 750 km/hr, -1000 is neglected.

{tex}2 \mathrm { x } ^ { 3 } + \mathrm { x } ^ { 2 } - 5 \mathrm { x } + 2 ; \frac { 1 } { 2 } , 1 , - 2{/tex}
Comparing the given polynomial with 
{tex}a x ^ { 3 } + b x ^ { 2 } + c x + d{/tex}, we get
a = 2, b = 1, c = 5, d = 2
Let {tex}p ( x ) = 2 x ^ { 3 } + x ^ { 2 } - 5 x + 2{/tex}
Then,
{tex}P \left( \frac { 1 } { 2 } \right) = 2 \left( \frac { 1 } { 2 } \right) ^ { 3 } + \left( \frac { 1 } { 2 } \right) ^ { 2 } - 5 \left( \frac { 1 } { 2 } \right) + 2{/tex}
{tex}= \frac { 1 } { 4 } + \frac { 1 } { 4 } - \frac { 5 } { 2 } + 2 = 0{/tex}
{tex}p ( 1 ) = 2 ( 1 ) ^ { 3 } + ( 1 ) ^ { 2 } - 5 ( 1 ) + 2{/tex}
= 2 + 1 - 5 + 2 = 0
{tex}p ( - 2 ) = 2 ( - 2 ) ^ { 3 } + ( - 2 ) ^ { 2 } - 5 ( - 2 ) + 2{/tex}
= -16 + 4 + 10 + 2 = 0
Therefore, {tex}\frac{1}{2}{/tex}, 1 and -2 are the zeroes of
{tex}2 x ^ { 3 } + x ^ { 2 } - 5 x + 2{/tex}
So, {tex}\alpha = \frac { 1 } { 2 } , \beta = 1 \text { and } \gamma = - 2{/tex}
Therefore,
{tex}\alpha + \beta + \gamma = \frac { 1 } { 2 } + 1 + ( - 2 ) = - \frac { 1 } { 2 } = - \frac { b } { a }{/tex}
{tex}\alpha \beta + \beta \gamma + \gamma \alpha = \left( \frac { 1 } { 2 } \right) \times ( 1 ) + ( 1 ) \times ( - 2 ) + ( - 2 ) \times \left( \frac { 1 } { 2 } \right){/tex}
{tex}= \frac { 1 } { 2 } - 2 - 1 = - \frac { 5 } { 2 } = \frac { c } { a }{/tex}
{tex}\alpha \beta \gamma = \left( \frac { 1 } { 2 } \right) \times ( 1 ) \times ( - 2 ) = - 1 = \frac { - 2 } { 2 } = \frac { - d } { a }{/tex}