Step:1 Since 420 > 130 we apply the division lemma to 420 and 130 to get ,
420 = 130 x 3 + 30
Step:2 Since 30 ≠ 0 , we apply the division lemma to 130 and 30 to get
130 = 30 x 4 + 10
Step:3 Since 10 ≠ 0 , we apply the division lemma to 30 and 10 to get
30 = 10 x 3 + 0
The remainder has now become zero, so our procedure stops. Since the divisor at this Step is 10, the HCF of 420 and 130 is 10.

An integer can be generated by multiplying the two smallest positive integers and it contains at least one divisor other than one and itself is called a composite number. Each positive <a href="https://byjus.com/maths/integers/">integers</a> is either prime, composite or the unit 1. Composite numbers are the numbers that are neither prime nor unit .
Example : 4( factors are 1, 2 and 4)

15 = 3×5
24 = 2×2×2×3
12 = 2×2×3
Lcm = 120
The greatest 4 digit no. Is 9999
Dividing lcm by 9999 we get remainder as 39
9999-39=9960
The answer is 9960

We have given that
4th term of an A.P.= a₄ = 0
∴ a + (4 – 1)d = 0
∴ a + 3d = 0
∴ a = –3d        ….(1)

25th term of an A.P. = a₂₅
= a + (25 – 1)d
= –3d + 24d      ….[From the equation (1)]
= 21d

3 times 11th term of an A.P. = 3a₁₁
= 3[a + (11 – 1)d]
= 3[a + 10d]
= 3[–3d + 10d]
= 3 × 7d
= 21d

∴ a₂₅ = 3a₁₁

i.e., the 25th term of the A.P. is three times its 11th term.

Let {tex}\alpha,\beta{/tex} be the zeros of the polynomial {tex}f(x)=x^2-8x+k{/tex}.
Sum of zeroes = {tex} \alpha + \beta = - \left( \frac { - 8 } { 1 } \right) = 8{/tex} and, Product of zeroes = {tex} \alpha \beta = \frac { k } { 1 } = k{/tex}
Now, 
{tex} \alpha ^ { 2 } + \beta ^ { 2 } = 40{/tex}

{tex} \Rightarrow \alpha ^ { 2 } + \beta ^ { 2 }+2 \alpha\beta-2 \alpha\beta= 40{/tex}
{tex} \Rightarrow \quad ( \alpha + \beta ) ^ { 2 } - 2 \alpha \beta = 40{/tex}
{tex} \Rightarrow \quad 8 ^ { 2 } - 2 k = 40{/tex}
{tex} \Rightarrow \quad 2 k = 64 - 40 {/tex}

{tex}\Rightarrow 2 k = 24 {/tex}

{tex}\Rightarrow k = 12{/tex}

Let a be the first term and d be the common difference of the given AP.

Now a₄=a+(4-1)d

{tex}\implies{/tex}a₄=a+3d.

And, a₁₇=a+(17-1)d

{tex}\implies{/tex}a₁₇= a+16d.
Then,
T₄=a+3d and T₁₇= a+16d
Now, {tex}\frac { T _ { 4 } } { T _ { 17 } } = \frac { 1 } { 5 }{/tex}
{tex}\Rightarrow \frac { a + 3 d } { a + 16 d } = \frac { 1 } { 5 }{/tex}
{tex}\Rightarrow{/tex} 5a+15d=a+16d

{tex}\implies{/tex}5a-a+15d-16d=0
{tex}\Rightarrow{/tex} 4a-d=0
{tex}\Rightarrow{/tex} 4a=d....(i)
Also, S₇=182

Where ,S₇={tex}\frac {7}2 [{2a+(7-1)d}]{/tex}
{tex}\Rightarrow \frac { 7 } { 2 } [ 2 a + 6 d ] = 182{/tex}
{tex}\Rightarrow \frac { 7 \times 2 } { 2 } [ a + 3 d ] = 182{/tex}
{tex}\Rightarrow{/tex} a+3d=26
{tex}\Rightarrow{/tex} a+3(4a)=26...[from (i)]
{tex}\Rightarrow{/tex} 13a=26
{tex}\Rightarrow{/tex} a=2
{tex}\Rightarrow{/tex} d=4(2)=8
Thus, We have
T₁ = 2
T₂ = T₁ + d = 2 + 8 = 10
T₃ = T₁ + 2d = 2 + 2(8) = 2 + 16 = 18
T₄ = T₁ + 3d = 2 + 3(8) = 2 + 24 = 26
Thus, the given AP is 2,10,18,26,...

If g(x)=x² + 2x + k is a factor of  f(x) = 2x⁴ + x³ - 14x² + 5x + 6

Then remainder is zero when f(x) is divided by g(x)

Let quotient =Q and remainder =R
Let us now divide f(x) by gx)

R = x(7k + 21) + (2k² + 8k + 6) -------(1)

and  Q = 2x² - 3x - 2(k + 4).------------(2)
{tex}\Rightarrow{/tex}x (7k + 21) + 2 (k² + 4k + 3) = 0 
{tex}\Rightarrow{/tex}7k + 21 = 0 and k² + 4k + 3 = 0
{tex}\Rightarrow{/tex} 7(k + 3) = 0 and (k + 1) (k + 3) = 0
{tex}\Rightarrow{/tex} k + 3 = 0
{tex}\Rightarrow{/tex}k = -3
Substituting the value of k in the divider x² + 2x + k, we obtain: x² + 2x - 3 = (x + 3) (x - 1) as the divisor.
Hence two zeros of g(x) are -3 and 1.------(3)
Putting k=-3 in (2) we get
Q = 2x² - 3x - 2
= 2x² - 4x + x - 2
= 2x(x - 2) + 1 (x - 2)
= (x - 2)(2x + 1)

Q=0 if x-2=0 or 2x+1=0
So other two zeros of f(x) are 2 and -{tex}\frac12{/tex}-------(4)

As g(x) is a factor of f(x) so zeros of G(x) are zeros of f(x) also

Hence from (3) and (4) we get
The zeros of f(x) are: -3 ,1,, 2 and  {tex}\frac12{/tex}

We will find zeroes of p(x)  which are not the zeroes of q(x) by using Factor theorem and Euclid’s division algorithm.

By factor theorem if q(x) is a factor of p(x), then r(x) must be zero.

p(x) = x⁵ – x⁴ – 4x³ + 3x² + 3x + b

q(x) = x³ + 2x² + a

So, by factor theorem remainder must be zero i.e.,

r(x) = 0
Value of r(x) is - ax² - x² + 3ax + 3x - 2a + b .
- ax² - x² + 3ax + 3x - 2a + b = 0x² + 0x + 0
⇒ -(a + 1)x² + (3a + 3)x + (b – 2a) = 0x² + 0x + 0

Comparing the coefficients of x², x and constant. on both sides, we get

-(a + 1) = 0 and 3a + 3 = 0 and b – 2a = 0
a + 1 = 0
a = -1
and  3a + 3 = 0
3a = - 3
a = -1
Put a = -1 in b - 2a = 0
Then b – 2(-1) = 0
⇒ b + 2 = 0
⇒ b = -2

For a = -1 and b = -2, zeroes of q(x) will be zeroes of p(x).

For zeroes of p(x),  p(x) = 0

⇒ (x³ + 2x² + a)(x² – 3x + 2) = 0 [∵ a = -1]

⇒ [x³ + 2x² – 1][x² – 2x – 1x + 2] =0

⇒ (x³ + 2x² – 1)[x(x – 2) – 1(x – 2) = 0

⇒ (x³ + 2x² – 1) (x – 2) (x – 1) = 0
⇒   (x – 2)  = 0 and (x – 1) = 0
⇒  x = 2 and x = 1

Hence, x = 2 and 1 are not the zeroes of q(x).

We need to find the H.C.F. of 506 and 1155 and express it as a linear combination of 506 and 1155.
By applying Euclid’s division lemma
1155 = 506 {tex}\times{/tex} 2 + 143.
506 = 143 {tex}\times{/tex} 3 + 77.
143 = 77 {tex}\times{/tex} 1 + 66.
77 = 66 {tex}\times{/tex} 1 + 11.
66 = 11 {tex}\times{/tex} 6 + 0.
Therefore, H.C.F. = 11.
Now, 11 = 77 - 66 {tex}\times{/tex} 1 = 77 - [143 - 77 {tex}\times{/tex} 1] {tex}\times{/tex} 1           {∵ 143 = 77 {tex}\times{/tex} 1 + 66}
= 77 - 143 {tex}\times{/tex} 1 + 77 {tex}\times{/tex} 1
= 77 {tex}\times{/tex} 2 - 143 {tex}\times{/tex} 1
= [506 - 143 {tex}\times{/tex} 3] {tex}\times{/tex} 2 - 143 {tex}\times{/tex} 1  {∵ 506 = 143 {tex}\times{/tex} 3 + 77 }
= 506 {tex}\times{/tex} 2 - 143 {tex}\times{/tex} 6 - 143 {tex}\times{/tex} 1
= 506 {tex}\times{/tex} 2 - 143 {tex}\times{/tex} 7
= 506 {tex}\times{/tex} 2 - [1155 - 506 {tex}\times{/tex} 2] {tex}\times{/tex} 7 {∵1155 = 506 {tex}\times{/tex} 2 + 143 }
= 506 {tex}\times{/tex} 2 - 1155 {tex}\times{/tex} 7 + 506 {tex}\times{/tex} 14
= 506 {tex}\times{/tex} 16 - 1155 {tex}\times{/tex} 7
Hence obtained.

For no solution,
{tex}\frac { a _ { 1 } } { a _ { 2 } } = \frac { b _ { 1 } } { b _ { 2 } } \neq \frac { c _ { 1 } } { c _ { 2 } }{/tex} {tex}\Rightarrow{/tex} {tex}\frac{k}{5}{/tex} = {tex}\frac{2}{-3}{/tex} {tex}\ne{/tex} {tex}\frac{-1}{2}{/tex}
{tex}\Rightarrow{/tex} k = {tex}\frac{-10}{3}{/tex}