Question

Factorise it ...........

............​

Answer 1

1.

2x² + 3√5x + 5

= 2x² + 2√5x + √5x + 5

= 2x(x + √5) + √5(x + √5)

= (x + √5) (2x + √5) <= ans

2.

The constant term in f(x) is are 1 and 2

putting x = 1 in f(x), we have

f(1) = (1)3 - 2(1)2 -1 + 2

= 1 - 2 - 1 + 2 = 0

According to remainder theorem f(1) = 0

so that,

(x - 1) is a factor of x3 - 2x2 - x + 2

Putting x = - 1 in f(x),

we have

f(-1) = (-1)3 - 2(-1)2

(-1) + 2

= -1 - 2 + 1 + 2 = 0

According to remainder theorem

f(-1) = 0

so that,

(x + 1) is a factor of x3 - 2x2 - x + 2

Putting x =2 in f(x),

we have

f(2) = (2)3 - 2(2)2

(2) + 2

= 8 -82 - 2 + 2 = 0

According to remainder theorem f(2) = 0

so that, (x = 2 ) is a factor of

x3 - 2x2 - x + 2

Here maximum power of x is 3 so that its can have maximum 3 factors

Here maximum power of x is 3 so that its can have maximum 3 factors So our answer is (x-1)(x+1)(x-2)

Answer 2

Heyya

Refer the pic .....