(i) Here you can see that
49a² = (7a)², 25b² = (5b)2, 70ab = 2(7a) (5b)
Comparing the given expression with x2 + 2xy + y2, we observe that x = 7a and y = 5b.
Using Identity I, we get
49a² + 70ab + 25b² = (7a + 5b)² = (7a + 5b) (7a + 5b)
(ii) We have \(\frac{25}{4} x^{2}-\frac{y^{2}}{9}=\left(\frac{5}{2} x\right)^{2}-\left(\frac{y}{3}\right)^{2}\)
Now comparing it with Identity III, we get
\(\frac{25}{4} x^{2}-\frac{y^{2}}{9}=\left(\frac{5}{2} x\right)^{2}-\left(\frac{y}{3}\right)^{2}\)
\(=\left(\frac{5}{2} x+\frac{y}{3}\right)\left(\frac{5}{2} x-\frac{y}{3}\right)\)

\(13^{\frac{1}{5}} \cdot 17^{\frac{1}{5}}=(13 \times 17)^{\frac{1}{5}}=221^{\frac{1}{5}}\)

Comparing the given expression with (x + y + z)², we find that
x = 3a, y = 4b and z = 5c.
Therefore, using Identity V, we have
(3a + 4b + 5c)² = (3a)² + (4b)² + (5c)² + 2(3a)(4b) + 2(4b)(5c) + 2(5c)(3a)
= 9a² + 16b² + 25c² + 24ab + 40bc + 30ac

As x – 1 is a factor of p(x) = 4x³ + 3x² – 4x + k, p(1) = 0
Now, p(1) = 4(1)³ + 3(1)² – 4(1) + k
So, 4 + 3 – 4 + k = 0
i.e., k = –3

28

3

\(2-\sqrt { 3 } \)

\(93-24\sqrt { 15 } \)

7/6,4/3,3/2,5/3 and 11/6

(i) x³ - 2x² - x + 2
Rearranging the terms, we have
x³ - 2x² - x + 2 = x³ - x - 2x² + 2
= x(x² - 1) - 2(x² - 1)
= [(x)² - (1)²][x - 2]
= (x - 1)(x + 1)(x - 2)
[∴ a² - b² = (a + b)(a - b)]
Thus,  x³ - 2x² - x + 2 = (x - 1)(x + 1)(x - 2)
(ii) Let p(x) = x³ - 3x² - 9x - 5
We shall find a factor of p(y) by using some trial value of y, say x = -1
p(-1) = (-1)³ - 3(-1)² - 9(-1) - 5
= -1 - 3 + 9 - 5
= -9 + 9 = 0
Since the remainder of p(-1) = 0 , by factor theorem we can say x + 1 is a factor ofp(x) = x³ - 3x² - 9x - 5.
Now dividing p(x) by x + 1 using long division.

Hence x³ - 3x² - 9x - 5 = (x +1) (x² - 4x - 5)
Now taking x² - 4x - 5, find two numbers p, q such that:
p + q = -4 (co-efficient of x)
p x q = 1 × -5 = -5 (product of co-efficient of x² and the constant term)
By trial and error method, we get p = -5, q = 1.
Now splitting the middle term of the given polynomial,
x² - 4x - 5 = x² - 5x + x - 5
= x(x - 5) + 1(x - 5)
= (x + 1)(x - 5)
(iii) x³ + 13x² + 32x + 20
We have p(x) = x³ + 13x² + 32x + 20
By trial, let us find: p(1) = (1)³ + 13(1)² + 32(1) + 20
= 1 + 13 + 32 + 20
= 66 ≠ 0
Now p(-1) = (-1)³ + 13(-1)²+ 32(-1) + 20
= -1 + 13 - 32 + 20
= 0
∴ By factor theorem, [x - (-1)], i.e. (x + 1) is a factor p(x).
∴ \(\frac{x^{3}-13 x^{2}-32 x-20}{(x+1)}\) = x² + 12x+ 20
or x³ + 13x² + 32x + 20 = (x + 1)(x² + 12x + 20)
= (x + 1)[x² + 2x + 10x + 20]
[Splitting the middle term]
= (x + 1)[x(x + 2) + 10(x + 2)]
= (x + 1)[(x + 2)(x + 10)]
= (x + 1)(x + 2)(x + 10)
(iv) 2y³ + y² - 2y - 1
We have p(y) = 2y³ + y² - 2y - 1
By trial, we have p(1) = 2(1)³ + (1)² - 2(1) - 1
= 2( 1) + 1 - 2 - 1
= 2 + 1 - 2 - 1 = 0
∴ By factor theorem, (y - 1) is a factor of p(y).
∴ \(\frac{\left(2 y^{3}+y^{2}-2 y-1\right)}{(y-1)}\) = 2y² + 3y + 1
∴ 2y³- y² - 2y - 1 = (y - 1)(2y² + 3y + 1)
= (y - 1)[2y² + 2y + y + 1]
[Splitting the middle term]
= (y - 1)[2y(y + 1) + 1(y + 1)]
= (y - 1)[(y + 1)(2y + 1)]
= (y - 1)(y + 1)(2y + 1)

(i) 2 + x² + x
The co-efficient of x² is 1.
(ii) 2 - x² + x3
The co-efficient of x² is (-1).
⁽ⁱⁱⁱ⁾ \(\frac{\pi}{2}\)x²+ x
The co-efficient of x² is \(\frac{\pi}{2}\).
(iv) \(\sqrt{2}\)x-1
∵  \(\sqrt{2}\)x-1 ⇒  \(\sqrt{2}\)x-1 + 0・x²
∴ The co-efficient of x² is 0.

\(-\frac { 31 }{ 4 } \)

(i) (9)^3/2 = (3²)^3/2=(3)³ = 27   [\(\because\) (a^m)ⁿ=(a)^mn]
(ii) (32)^2/5 = (2⁵)^2/5 = (2)² = 4
(iii) (16)^3/4 = (2⁴)^3/4 = (2)³ = 8
(iv) (125)^-1/3 = \(\frac { 1 }{ { (125 })^{ 1/3 } } =\frac { 1 }{ ({ 5 }^{ 3 })^{ 1/3 } } =\frac { 1 }{ 5 } \)
\(\left[ \because \ { (a) }^{ -n }=\frac { 1 }{ { a }^{ n } } and\ ({ a }^{ m })^{ n }=({ a) }^{ mn } \right] \)

We have, \(\sqrt { 5 } \)
Now, 5 can be written as 5 = 4 + 1 = (2)² + (1)²
So, we take OQ = 2 and PQ = 1
Now, draw OQ = 2 units on the number line and then draw PQ=1 unit, such that \(OP\bot OQ\) join OP

Now, by Pythagoras theorem, we get
OP² = OQ² + PQ²
On taking positive square root, we get
\(OP=\sqrt { { OQ }^{ 2 }+{ PQ }^{ 2 } } =\sqrt { { 2 }^{ 2 }+{ 1 }^{ 2 } } \)
\(=\sqrt { 4+1 } =\sqrt { 5 } \)
On taking O as centre and Op = \(\sqrt { 5 } \)  as radius, draw an arc, which intersects the number line at point R.
Hence, the point R represents \(\sqrt { 5 } \)