Question

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Need solutions​

Answer 1

3)[A] x(x+y) - 10x - 10y

= x(x+y)-10(x+y)

= (x+y) (x-10)

[B] 16x² + 40xy + 25y²

= (4x)² + (5y)² + 2.4x.5y

= (4x+5y)²

[C] m⁴ - n⁴

= (m²)² - (n²)²

= (m²+n²) (m²-n²)

= (m+n) (m-n) (m²+n²)

[D] a² - 7a + 12

= a²-(4+3)a+12

=a² - 4a - 3a + 12

= a(a-4)-3(a-4)

= (a-3) (a-4)

[E] x²+y²-z²+2xy

= (x+y)² - (z)²

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

[F] m²-m-6

= m²-(3-2)m-6

= m²-3m+2m-6

= m(m-3)+2(m-3)

= (m-3) (m+2)

Answer 2

a. x( x + y ) - 10x - 10y

= > x( x + y ) - 10x - 10y

= > x( x + y ) - 10( x + y )

= > ( x + y )( x - 10 )

Thus x( x + y ) - 10x - 10y is equal to ( x + y )( x - 10 ), when factorised.

b. 16x^2 + 40xy + 25y^2

= > 16x^2 + 40xy + 25y^2

= > ( 4x )^2 + 2( 4x × 5y ) + ( 5y )^2

= > ( 4x + 5y )^2 { formula : a^2 + 2ab + b^2 = ( a + b )^2 }

= > ( 4x + 5y )( 4x + 5y )

Thus, 16x^2 + 40xy + 25y^2 is equal to ( 4x + 5y )( 4x + 5y ), when factorised.

c. m^4 - n^4

= > m^4 - n^4

= > ( m^2 )^2 - ( n^2 )^2

= > ( m^2 - n^2 )( m^2 + n^2 ) { by using a^2 - b^2 = ( a + b )( a - b ) }

= > ( m + n )( m - n )( m^2 + n^2 ) { by using the above formula }

Thus m^4 - n^4 is equal to ( m + n )( m - n )( m^2 + n^2 ).

d. a^2 - 7a + 12

= > a^2 - 7a + 12

= > a^2 - ( 4 + 3 )a + 12

= > a^2 - 4a - 3a + 12

= > a( a - 4 ) - 3( a - 4 )

= > ( a - 3 )( a - 4 )

Thus a^2 - 7a + 12 is equal to ( a - 3 )( a - 4 ), when factorised.

e. x^2 + y^2 - z^2 + 2xy

= > x^2 + y^2 + 2xy - z^2

= > ( x + y )^2 - z^2 { by using a^2 + b^2 + 2ab = ( a + b )^2 }

= > ( x + y + z )( x + y - z )

f. m^2 - m - 6

= > m^2 - 3m + 2m - 6

= > m( m - 3 ) + 2( m - 3 )

= > ( m - 3 )( m + 2 )