Question

Factorize :-
i)
$(a + b)^{2} - 11(a + b) - 42$
ii)
$7 + 10(2x - 3y) - 8(2x - 3y)^{2}$
Please send the whole method


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Answer 1

We Have :-

( a + b )² - 11 ( a + b ) - 42

7 + 10 ( 2x - 3y ) - 8 ( 2x - 3y )²

To Do :-

Factorize

Method Used :-

Middle Term Splitting

Solution :-

( i ) ( a + b )² - 11 ( a + b ) - 42

Let ( a + b ) = n

( n )² - 11 ( n ) - 42

n² - 14n + 3n - 42

n ( n - 14 ) + 3 ( n - 14 )

( n - 14 ) ( n + 3 )

( a + b - 14 ) ( a + b + 3 )

( ii ) 7 + 10 ( 2x - 3y ) - 8 ( 2x - 3y )²

Let ( 2x - 3y ) = m

7 + 10 ( m ) - 8 ( m )²

- 8m² + 10m + 7

8m² - 10m - 7

8m² - 14m + 4m - 7

2m ( 4m - 7 ) + 1 ( 4m - 7 )

( 4m - 7 ) ( 2m + 1 )

[ 4 ( 2x - 3y ) - 7 ] [ 2 ( 2x - 3y ) + 1 ]

( 8x - 12y - 7 ) ( 4x - 6y + 1 )

Answer 2

Question :--

Factorize :------

• (a+b)² - 11(a+b) - 42
• 7 + 10(2x-3y) - 8(2x-3y)²

Solution :--

Answer (1)

→ (a+b)² - 11(a+b) - 42

Taking (a+b) common , we get,

→ (a+b) [ (a+b) - 11 - 42) ]

→ (a+b) [ a+b - 53 ]

→ (a+b)(a+b - 53) This is simplest form we can factorize .

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Answer 2 :--

→ 7 + 10(2x-3y) - 8(2x-3y)²

→ (-8)(2x-3y)² + 10(2x-3y) + 7

Now, Either we take (2x-3y) common From here only, as its not a big polynomial, or to make it little bit easy lets assume (2x-3y) as R .

So , putting this we get,

→ (-8)R² + 10R + 7

Now, for Solving any Quadratic Equation , first we have to Put that Equal to zero.

→ (-8)R² + 10R + 7 = 0

Now , using Splitting the middle term Method , we get,

→ (-8)R² - 4R + 14R + 7 = 0

→ (-4R)[2R +1 ] + 7[2R+1] = 0

→ [2R+1] [ 7 -4R] = 0

Now, putting our value R = (2x-3y) we get,

→ [ 2(2x-3y) + 1 ] [ 7 - 4(2x-3y) ]

→ [ 4x - 6y + 1 ] [ 7 - 8x + 12y ]

→ [ 4x - 6y + 1 ] [12y - 8x + 7 ]

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