**Factorize the following:**

**(i) 9x**^{2}**– 16y**^{2}

Solution:

= (3x)^{2} – (4y)^{2}

Using a^{2} – b^{2} = (a + b) (a – b)

= (3x + 4y) (3x – 4y)

**(ii) 5x**^{2}**– 7y**^{2}**[ Hint : ( ****𝟓 ****)**^{2}**= 5 ] **

Solution:

( 5 x)^{2} – ( 7 y)^{2}

Using a^{2} – b^{2} = (a + b) (a – b)

= ( 5 x + 7 y) ( 5 x – 7 y)

**(iii) 100 – 9x****2 **

Solution:

= 102 – (3x)2

= (10 + 3x) (10 – 3x) [∵a^{2} – b^{2} = (a + b) (a – b)]

**(iv) (x + 4y)**^{2}**– 4z**^{2}

Solution:

= (x + 4y)^{2} – (2z)^{2} [∵a^{2} – b^{2} = (a + b) (a – b)]

= (x + 4y + 2z) (x + 4y – 2z)

**(v) x – 64xy**^{4}

Solution:

= x (1 – 64y^{4})

= x (1 + (8y^{2})^{2}) [∵a^{2} – b^{2} = (a + b) (a – b)]

= x (1 + 8y^{2}) (1 – 8y^{2})

**(vi) a**^{2}**+ 2ab + b**^{2}**– 9c**^{2}

Solution:

= (a + b)^{2} – (3c)^{2} [∵(a + b)^{2} = a^{2} + 2ab + b^{2} ;a^{2} – b^{2} = (a + b) (a – b)]

= (a + b + 3c) (a + b – 3c)

**(vii) 4x**^{2}**– 9y**^{2}**– 2x – 3y **

Solution:

= (2x)^{2} – (3y)^{2} – (2x + 3y)

= [(2x + 3y) (2x – 3y)] – (2x – 3y) [∵a^{2} – b^{2 }= (a + b) (a – b)]

= (2x + 3y) (2x – 3y – 1)

**(viii) a**^{2}**+ b**^{2}**– a + b **

Solution:

= a^{2} + b^{2 }– (a – b) [∵a^{2} – b^{2} = (a + b) (a – b)]

= (a + b) (a – b) – (a – b)

= (a – b) (a + b – 1)

**(ix) x**^{4}**– 625 **

Solution:

= (x^{2})^{2} – 25^{2} [∵a^{2} – b^{2} = (a + b) (a – b)]

= (x + 25) (x^{2} – 25)

= (x^{2} + 25) (x^{2} – 5^{2})

= (x^{2} + 25) (x + 5) (x – 5)

**(x) 36 (5x + y)**^{2}**– 25 (4x – y)**^{2 }

Solution:

= [6 (5x + y)]^{2} – [5 (4x – y)]^{2} [∵a^{2 }– b^{2} = (a + b) (a – b)]

= (30x + 6y + 20x – 5y) (30x + 6y – 20x + 5y)

= (50x + y) (10x + 11y)

**(xi) (a + **^{𝟏}**/ _{𝐚} **

**)**

^{2}**– 4 (x –**

^{𝟏}**/**

_{𝐱}**)**

^{2}Solution:

= (a + ^{𝟏}/_{𝐚} )^{2} – [2 (x – ^{𝟏}/_{𝐱} )]^{2}

= [a+ ^{𝟏}/_{𝐚} + 2( x− ^{𝟏}/_{𝐱})][ a+ ^{𝟏}/_{𝐚} – 2(x− ^{𝟏}/_{𝐱} )]

**(xii) 12m**^{2}**– 75n**^{2}

Solution:

= ( √12 m^{2})^{2} – (√75 n^{2})^{2} [∵a^{2} – b^{2} = (a + b) (a – b)]

= (2 √3m^{2})^{2} – (5√3 n^{2})^{2 }[ 12 =2√3 and 75 = 5√3

= (2√3m^{2} + 5√3 n^{2}) (2√3 m^{2} – 5√3 n^{2})

= √3 (2m^{2} + 5n^{2}) (2m – 5n) (2m + 5n)

**Factorize by adding and subtracting appropriate quantity.**

**(i) x**^{2}**+ 6x + 3 **

Solution:

= x^{2} + ^{2}. x. 3 + 3^{2} – 3^{2} + 3

= (x + 3)^{2} – 9 + 3

= (x + 3)^{2} – 6 [∵a^{2} – b^{2} = (a + b) (a – b)]

= (x + 3)^{2} – 6^{2}

= (x + 3 + 6 ) (x + 3 – 6 )

**(ii) x**^{2}**+ 10x + 8 **

Solution:

= x^{2} + 2. x. 5 + 5^{2} – 5^{2} + 8

= (x + 5)^{2} – 25 + 8

= (x + 5)^{2} – 17 [∵a^{2} + 2ab + b^{2} = (a + b)^{2}]

= (x + 5)^{2} – 17^{2}

= (x + 5 + 17 ) (x + 5 – 17 )

**(iii) x**^{2}**+ 10x + 20 **

Solution:

= x^{2} + 2. x. 5 + 5^{2} – 5^{2} + 20 [∵a^{2} + 2ab + b^{2} = (a + b)^{2}]

= (x + 5)^{2} – 25 + 20

= (x + 5)^{2} – 5

= (x + 5)^{2} – 5^{2} [∵a^{2} – b^{2} = (a + b) (a – b)]

= (x + 5 + 5 ) (x + 5 – 5 )

**(iv) x**^{2}**+ 2x – 1 **

Solution:

= x^{2} + 2. x. 1 + 1^{2} – 1^{2} + 1 [∵a^{2} + 2ab + b^{2} = (a + b)^{2}]

= (x + 1)^{2} – 1 + 1

= (x + 1)^{2} – 2

= (x + 1)^{2} – 2^{2} [∵a^{2} – b^{2 }= (a + b) (a – b)]

= (x + 1 + 2 ) (x + 1 – 2 )

**prove that (1 +**^{𝐱}**/**_{𝐲}**) (1 –**^{𝐱}**/**_{𝐲}**) (1 +****𝐱**^{𝟐}**/****𝐲**^{𝟐}**) (1 +****𝐱**^{𝟒}**/****𝐲**^{𝟒}**) +****𝐱**^{𝟖}**/𝐲**^{𝟖}**= 1**

Solution:

L.H.S = (1 +^{ 𝐱}/_{𝐲} ) (1 – ^{𝐱}/_{𝐲} ) (1 + 𝐱^{𝟐}/𝐲^{𝟐}) (1 + 𝐱^{𝟒}/𝐲^{𝟒} ) + 𝐱^{𝟖}/𝐲^{𝟖}

[∵a^{2} – b^{2} = (a + b) (a – b)]

= (1 – 𝐱^{𝟐}/𝐲^{𝟐} ) (1 + 𝐱^{𝟐}/𝐲^{𝟐} ) (1 + 𝐱^{𝟒}/𝐲^{𝟒}) + 𝐱^{𝟖}/𝐲^{𝟖}

= (1 – 𝐱^{𝟒}/𝐲^{𝟒}) (1 + 𝐱^{𝟒}/𝐲^{𝟒} ) + 𝐱^{𝟖}/𝐲^{𝟖}

= (1)^{2} – (𝐱^{𝟒}/𝐲^{𝟒})^{2} + 𝐱^{𝟖}/𝐲^{𝟖}

= 1 – 𝐱^{𝟖}/𝐲^{𝟖} + 𝐱^{𝟖}/𝐲^{𝟖}

= 1 R.H.S

**If x****=**^{𝐚+𝐛}**/**_{𝐚−𝐛}**and y****=**^{𝐚−𝐛}**/**_{𝐚+𝐛}**find x**^{2}**– y**^{2}**.**

Solution:

x^{2} – y^{2} = (x + y) (x – y)

= (^{a+b}/_{a−b}+^{a−b}/_{a+b})(^{a+b}/_{a−b }– ^{a−b}/_{a+b})

= [[(a+b)^{2}+ (a−b)^{2} ]/(a^{2}− b^{2})][[(a+b)^{2}− (a−b)^{2}]/(a^{2}− b^{2})]

= {(a^{2}+ 2ab + b^{2}+ a^{2}+ b^{2}− 2ab)/[(a^{2}− b^{2})^{2}]} x (a^{2} + 2ab + b^{2} – a^{2} – b2 + 2ab)

= [(2a^{2}+ 2b^{2})4ab]/[(a^{2}− b^{2})^{2}]

= [2(a^{2}+ b^{2})4ab]/(a^{2}− b^{2})^{2}

= [8ab(a^{2}+ b^{2})]/(a^{2}− b^{2})^{2}

**If p = x – y, q = y – z and r = z – x, simplify r**^{2}**– p + 2pq – q**^{2}

Solution:

(z – x)^{2} – (x – y)^{2} + 2 (x – y) (y – z) – (y – z)^{2}

= z^{2} – 2xy + x^{2} – (x^{2} – 2xy + y^{2}) + 2 (xy – y^{2} – xz + yz) – (y^{2} + z^{2} – 2yz)

= z^{2} + x^{2} – 2xz – x^{2} + 2xy – y^{2} + 2xy – 2y^{2} – 2xz + 2yz – y^{2} – z^{2} + 2yz

= 4xy – 4xz + 4yz – 4y^{2}

= 4[xy – xz + yz – y^{2}]

= 4[x (y – z) – y (y – z)]

= 4 (x – y) (y – z)

= 4pq

**The radius of a circle is 13cm in which a chord of 10cm is drawn. Find the distance of the chord from the centre of the circle.**

**[Hint: use Pythagoras rule and the fact that the perpendicular line drawn from the centre of the circle to any o the chords bisects the chord] **

Solution:

Let AC = 10cm is the chord of the circle with radius 13cm then using Pythagoras theory

OA^{2} = OB^{2} + AB^{2}

13^{2} = x^{2} + 5^{2}

x^{2} = 13^{2} – 5^{2}

= (13 + 5) (13 – 5)

= 18 (8)

x^{2} = 144 (∵ the perpendicular line drawn from the centre of the circle to a chord bisects the chord)

x = 144

x = 12 cm

**Is it possible to factorize x**^{2}**+ 4x + 20? Give reason**

Solution:

x^{2} + 4x + 20

= x^{2} + 2. x. 2 + 2^{2} – 2^{2} + 20

= x^{2} + 4x + 4 – 4 + 20

= (x + 4)^{2} + 16

= (x + 4)^{2} + 4^{2}

We cannot factorize their further since this is nit of the form a^{2} – b^{2}

Fantastic

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