Question

(2x-4)^2-11(2x-4)+28 factorii by steps plzz

Answer 1

Answer: Let (2x-4) = t

Now, factorise:

t^2- 11t+28 = t^2-7t- 4t +28

= t(t-7)-4(t-7)

= (t-7)(t-4)

Now in place of 't', put 2x-4,

(t-7)(t-4) = (2x-4-7)(2x-4-4)

= (2x-11)(2x-8)

Answer 2

Qᴜᴇsᴛɪᴏɴ :-

• Factorise :- (2x - 4)² - 11(2x - 4) + 28 ?

Sᴏʟᴜᴛɪᴏɴ :-

⟿ (2x - 4)² - 11(2x - 4) + 28

⟿ 2²(x - 2)² - 22x + 44 + 28

⟿ 4(x - 2)² - 22x + 72

using (a - b)² = a² + b² - 2ab Now,

⟿ 4(x² + 4 - 4x) - 22x + 72

⟿ 4x² - 16x + 16 - 22x + 72

⟿ 4x² - 16x - 22x + 16 + 72

⟿ 4x² - 38x + 88

⟿ 2(2x² - 19x + 44)

Splitting The Middle - Term Now,

⟿ 2(2x² - 8x - 11x + 44)

⟿ 2[ 2x(x - 4) - 11(x - 4) ]

⟿ 2[ (x - 4)(2x - 11) ]

⟿ 2(x - 4)(2x - 11) (Ans.)