🙂

If you have an expression:

ax^2 + bx + c

to factor into the parts

(dx+e)(fx+g)

then it should be clear from multiplying out the two terms above,

into

df x^2 + (dg+ef)x + eg

Since the mutipliers for x^2 in your original expression and this longer form must match, then

a = df

and also

b = (dg+ef)

and also

c = eg

Look at the expression you started with, and you can see

that a = 1; b = 7; c = 6

Often, simple choices make the solution simple.

Since a = 1, therefore df = 1;

The simplest choice for d and f is 1 for each.

Make that guess so that the next steps are simple.

If c = 6, then eg = 6. The simplest guesses for e & g are:

1 & 7 and

2 & 6. Try each pair so that you can arrive at d and f.

If you can find values for d and f that enable

dg+ef to equal 7, then you have solved for the values you need to plug into that two part set of parenthesized factors I described as

(dx+e)(fx+g).

Let’s plug in the assumed values of (1) for each of d & f:

(x+e)(x+g)

Now plug in values for e & g, a pair at a time to see what multiplies out as required.

(x+1)(x+7) gives what?

(x+2)(x+3) gives what?

Which substitution gives the desired result?

( ) ( )

2. since its x squared put x in front of each bracket.

(x ) (x )

add b term (7x) 2 things that add to 7, multiply c (6) the 2 numbers that add up to 7 have to multiple to be 6, so by trial and error the numbers are: 6, 1(6 + 1=7, 6×1 = 6)

therefore the answer is:

(x +1) (x+6)

hope this helps you.