Question

factorise (x-3)^2-7(x-3)-10​

Answer 1

Corrected Question:-

Factorise (x-3)²-7(x-3)+10

Solution :-

Given expression is (x-3)²-7(x-3)+10

=> (x-3)²-2(x-3)-5(x-3)+10

=> [(x-3){(x-3)-2}]-5[(x-3)-2]

=> [(x-3)-2](x-3-5)

=> (x-3-2)(x-3-5)

=> (x-5)(x-8)

Alternative Method:-

Given expression is (x-3)²-7(x-3)+10

Put x-3 = a then

=> a²-7a+10

=> a²-2a-5a+10

=> a(a-2)-5(a-2)

=> (a-2)(a-5)

Now,

=> (x-3-2)(x-3-5)

=> (x-5)(x-8)

Answer :-

(x-3)²-7(x-3)+10 = (x-5)(x-8)

Used Method:-

→ Splitting the middle term

Check:-

(x-3)²-7(x-3)+10

=> x²-6x+9-7x+21+10

=> x²-13x+40

=> x²-5x-8x+40

=> x(x-5)-8(x-5)

=> (x-5)(x-8)

Verified the given relations in the given problem.

Answer 2

Given :-

(x - 3)² - 7(x - 3) - 10 = 0

To Find :-

Factorized form

Solution :-

We know that

(a - b)² = a² + b² - 2ab

$\rm\dashrightarrow (x-3)^2-7(x-3)-10=0$

$\rm\dashrightarrow x^2+(3)^2-2(3)(x)-7(x-3)-10=0$

$\rm\dashrightarrow x^2+9-6x-7x+21-10=0$

$\rm\dashrightarrow x^2-13x+20=0$

It can't be factorized further