Question

Factorise using identity x²+6x+7​

Answer 1

Answer: x^2-6x-7= (x+1) (x-7)

Step-by-step explanation:
(x+a) (c+b) =x^2 + (a +b)x + ab
So if x^2- 6x-7 has factors of the form
(x + a)(x + b) then a +b= - 6 and ab =-7
So we are looking for factors a and b of
(- 7) which add up to ( - 6)
If we optimistically assume that a and b are integers, the only factors of ( - 7) are (a, b) = (- 1, 7) which gives a sum of ( + 6) and (a, b) = (1, - 7) which gives the sum we want: ( - 6)
So (x+ a) (x+b)= (x+1) (x-7)

Answer 2

To factorise the quadratic expression x² + 6x + 7, we need to find two binomials whose product equals the original expression.

One way to do this is to use the fact that (a + b)² = a² + 2ab + b², which can be rearranged to give:

a² + 2ab + b² = (a + b)²

Using this identity, we can rewrite the expression x² + 6x + 7 as:

x² + 6x + 7 = x² + 2(3)(x) + 3² - 3² + 7

Notice that we added and subtracted 3² = 9 inside the parentheses. Now we can use the identity above to write:

x² + 6x + 7 = (x + 3)² - 2² + 7

Simplifying the expression inside the parentheses gives:

x² + 6x + 7 = (x + 3)² - 4

Therefore, we have factored the quadratic expression x² + 6x + 7 as:

x² + 6x + 7 = (x + 3)² - 4