(i) x³ - 3x² -10x + 24
Let p(x) = x³ - 3x² -10x + 24
Sum of all the co-efficients = 1- 3 - 10 + 24 = 25 -13 = 12 ≠ 0
Hence (x - 1) is not a factor.
Sum of co-efficient of even powers with constant = -3 + 24 = 21
Sum of co-efficients of odd powers = 1 - 10 = - 9
21 ≠ -9
Hence (x + 1) is not a factor.
p(2) = 2³+ 3(2²) -10\(\times\)2+24
= 8 -12 - 20 + 24
= 32 - 32 = 0 ஃ (x - 2) is a factor.
Now we use synthetic division to find other factor

Thus (x - 2) (x + 3) (x - 4) are the factors.
ஃ x³ - 3x² - 10x + 24 = (x - 2)(x + 3)(x - 4)
(ii) 2x² - 3x² - 3x + 2
Let p (x) = 2x³ - 3x² - 3x + 2
Sum of all the co-efficients are
2 - 3 - 3 + 2 = 4 - 6 = -2 ≠ 0
ஃ (x - 1) is not a factor.
Sum of co-efficients of even powers of x with constant = -3 + 2 = - 1
Sum of co-efficients of odd powers of x = 2 - 3 = - 1
(-1) = (-1)
ஃ (x + 1) is a factor
Let us find the other factors using synthetic division

Quotient is 2x²- 5x + 2 = 2x - 4x - x + 2 = 2x (x - 2) - 1 (x - 2)
= (x - 2)(2x - 1)
ஃ 2x³- 3x²- 3x +2 = (x +1) (x - 2) (2x - 1)
(iii) -7x+ 3 + 4x³
Letp (x) = 4x³ + 0x² - 7x + 3
Sum of the co-efficients are = 4 + 0 - 7 + 3
= 7 - 7 = 0
ஃ (x - 1) is a factor
Sum of co-efficients of even powers of x with constant = 0 + 3 = 3
Sum of co-efficients of odd powers of x with constant = 4 - 7 = -3
3 ≠ -3
ஃ (x + 1) is not a factor
Using synthetic division, let us find the other factors.


Quotient is 4x² + 4x - 3
= 4x²+ 6x - 2x - 3
= 2x(2x + 3) -1(2x + 3)
= (2x + 3)(2x - 1)
ஃ The factors are (x -1), (2x + 3) and (2x - 1)
ஃ -7x + 3 + 4x³= (x+1)(2x+3)(2x-1)
(iv) x³ + x² - 14x - 24
Let p(x) = x³ + x² - 14x - 24
Sum of the co-efficients are 1 + 1-14 - 24 = -36 ≠ 0
ஃ (x - 1) is not a factor
Sum of co-efficients of even powers of x with constant = 1 - 24 = -23
Sum of co-efficients of odd powers of x = 1 - 14 = -3
-23 ≠ -13
ஃ (x + 1) is also not a factor
p(2) = 2³ + 2² - 14 (2) - 24 = 8 + 4 - 28 - 24
= 12 - 52 ≠ 0, (x - 2) is a not a factor
p(-2) = (-2)³ + (-2)² - 14 (-2) - 24
-8 + 4 + 28 - 24 = 32 - 32 = 0
ஃ (x + 2) is a factor
To find the other factors let us use synthetic division
x³+x²-14x-24


ஃ The factors are (x + 2),(x + 3),(x - 4)
ஃ x³+x²-14x - 24 = (x + 2)(x + 3)(x - 4)
(v) x³- 7x + 6
Let p (x) = x³+ 0x²-7x + 6
Sum of the co-efficients are = 1+0 -7+ 6 = 7-7 = 0
ஃ (x - 1) is a factor
Sum of co-efficients of even powers of x with constant = 0 + 6 = 6
Sum of coefficient of odd powers of x = 1 - 7 = - 7
6 ≠ -7
ஃ (x + 1) is not a factor
To find the other factors, let us use synthetic division.


ஃ The factors are (x - 1), (x - 2), (x + 3)
ஃ x³ + 0x² - 7x + 6 = (x - 1)(x - 2)(x + 3)
(vi) x³-10x²- x + 10
Let p(x) = x³-10x²- x + 10
Sum of the co-efficients = 1 - 0 - 1 + 10
= 11 -11 = 0
(x - 1) is a factor
Sum of co-efficients of even powers of x with constant = -10 + 10 = 0
Sum of co-efficients of odd powers of = 1 - 1 = 0
ஃ (x + 1) is a factor
Synthetic division

ஃ x³ + 10x²- x + 10 = (x - 1) (x + 1)(x -10)

Let p(x) = x −x³ − 5x²- 2x + 24
When x = 1, p(1)= 1− 5 − 2 + 24 = 18 ≠ 0  (x -1) is not a factor.
When x = –1, p(-1) = -1− 5 + 2 + 24 = 20 ≠ 0  (x + 1) is not a factor.
Therefore, we have to search for different values of x by trial and error method.
When x = 2
p(2) = 2x³ − 5(2)² − 2(2) + 24 
= 8 − 20 − 4 + 24
= 8 ≠ 0
Hence, (x – 2) is not a factor
When x = −2
p(-2) = (−2)³ −5(−2)² −2(−2) +24
= −8 −20 + 4 +24
p(-2) = 0
Hence, (x + 2) is a factor

Thus, (x + 2)(x − 3)(x − 4) are the factors.
Therefore, x³- 5x²- 2x²+ 24 = (x + 2)(x − 3)(x − 4)

(i) (x³+x²-7x-3) ÷ (x-3)
Let p (x) = x³+x²-7x-3
q (x) = x - 3
To find the zero of x - 3 :
p(x) in standard form ((i.e.) descending order)
Co-efficients are 1 1 -7 -3

Quotient is:  x² +4x+5
Remainder is :12
(ii) (x³+2x²-x-4) ÷ (x+2)
p(x) = x³+2x²-x-4
Co-efficients are 1 2 -1 -4
To find zero of x+2, put x+2 = 0; x = -2

Quotient is: (x²-1)
Remainder is: -2
(iii) (3x³-2x²+7x-5) ÷ (x+3)
To find zero of the divisor (x + 3), put x + 3 = 0;  x = - 3
Dividend in Standard form 3x³-2x²+7x-5
Co-efficients are 3 -2 7 -5
Synthetic Division

Quotient is: 3x²-11x+40
Remainder: is -125
(iv) (8x⁴-2x²+6x+5) ÷ (4x+1)
To find zero of the divisor 4x + 1, put 4x + 1 = 0; 4x = -1; x = \(-\frac{1}{4}\)
Dividend in Standard form 8x⁴-2x²+6x+5
Co-efficients are 8 0 -2 6 5
Synthetic Division

8x⁴-2x²+6x+5 = \((x+\frac{1}{4})(8x^3-2x^2-\frac{3x}{2}+\frac{51}{8})+\frac{109}{32}\)
=\(\frac{4x+1}{ ̶4̶}\times ̶4̶(2x^3-\frac{x^2}{2}-\frac{3x}{8}+\frac{51}{32})+\frac{109}{32}\)
= (4x+1)\((2x^3-\frac{x^2}{2}-\frac{3x}{8}+\frac{51}{32})+\frac{109}{32}\)
Quotient is: \(2x^2-{x^2\over2}-{3x\over8}+{51\over32}\)
Remainder is: \({109\over32}\)

(i) 2a²+ 9a +10 = 2a²+ 4a + 5a + 10
= 2a(a + 2)+ 5(a + 2)
= (a + 2)(2a + 5)

(ii) 5x²- 29xy - 42y²
= 5x²- 35xy + 6xy - 42y²
= 5x(x - 7y)+ 6y(x - 7y)
= (x - 7y)(5x - 6y)

(iii) 9 -18x + 8x²
= 8x²-18x + 9
= 8x²- 6x -12x + 9
= 4x(2x - 3) - 3(2x - 3)
= (4x - 3)(4x - 3)

(iv) 6x² + 16xy + 8y²
= 2(3x² + 8xy + 4y²)
= 2(3x² + 8xy + 4y²)
= 2(3x² + 6xy + 2xy + 4y²)
= 2(3x (x + 2y)- 2y(x + 2y))
= 2(x + 2y)(3x + 2y)

(v) 12x² + 36x²y + 27y²x²
= 27y²x²+36x²y+12x²= 3x²(9y²+12y+4)
= 3x²(9y²+6y+6y+4) = 3x²(3y(3y+2)+2(3y+2))
= 3x²(3y+2)(3y+2) = 3x²(3y+2)(3y+2)
= 3x²(3y + 2)²

(vi) (a + b)²+ 9(a + b) + 18
= (a + b)²+ 6(a + b) + 3(a + b) + 18
= (a + b)((a + b)+6)+3((a + b)+6)
= ((a + b)+6)((a + b)+3) = (a + b + 6)(a + b + 3)

Let x + y = p , we get p² + 9p + 20
Compare with ax² + bx + c
We get a = 1, b = 9, c = 20

The required factors are 4 and 5
product ac = 1 \(\times\) 20 = 20, sum b = 9
∴ we split the middle term as 4p and 5p
p² + 9p + 20 = p² + 4p + 5p + 20
= p (p+4)+ 5(p + 4)
= (p + 4) (p + 5)
Put, p = x + y we get, (x + y²)+9 (x + y) + 20 = (x + y + 4) (x + y + 5)

Compare with ax² + bx + c
Here, a = 2, b = 15, c = −27
product ac = 2\(\times\)–27 = –54, sum b = 15

The required factors are –3 and 18
∴ we split the middle term as 18x and –3x
2x² + 15x - 27 = 2x² +  18x − 3x - 27
= 2x (x + 9) − 3(x + 9)
= (x + 9) (2x − 3)
Therefore, (x + 9) (2x − 3) and are the factors of 2x² + 15x - 27

Compare with ax²+bx+c
we get, a = 2, b = -15, c = 27
product ac = 2\(\times\) 27 = 54 and sum b = -15
∴ we split the middle term as -6x and -9x

The required factors are -6 and -9
2x²-15x + 27 = 2x²- 6x - 9x + 27
=2x(x - 3) -9(x - 3)
=(x - 3)(2x - 9)
Therefore, (x - 3) and (2x - 9) are the factors of 2x²-15x + 27

\(\frac{1}{15}\)

Let \(\frac{(x^2-y^2)^3+(y^2-z^2)^3+(z^2-x^2)^3}{(x-y)^3+(y-z)^3+(z-x)^3}\) ...(1)

(x²-y²)³+(y²-z²)³+(z²-x²)³=3(x²-y²)(y²-z²)(z²-x²)
(x-y)³+(y-z)³+(z-x)³=3(x-y)(y-z)(z-x)

=(x+y)(y+z)(z+x)

(i) 98² = (100-2)³
(a-b)³ ≡ a³- 3a²b + 3ab²- b³
98³ = (100 - 2)² = 100³- 3\(\times\)100²\(\times\)2 + 3\(\times\)100\(\times\)2²- 2³
= 1000000 - 3\(\times\)10000\(\times\)2 + 300\(\times\)4 - 8
= 1000000 - 60000 +1200 - 8 =1001200 - 60008
= 941192
(ii) 1001³= (1000 + 1)³
(a+b)³ ≡ a³+3a²b+3ab²+b³
(1000+1)³=1000³+3(1000)\(\times\)1²+1³
= 1000,000,000 + 3,000,000 + 3000 + 1
= 1,003,003,001