Multiply out the brackets?

Question 1
x ² - 3 x - 6 x + 18
x ² - 9 x + 18

Question 2
( x + 3 ) ( x - 1 )
x ² + 2 x - 3

(x-6)(x-3) = x²-6x-3x+18 = x²-9x+18

(x+3)(x-1) = x²+3x-x-3 = x²+2x-3

Multiplying out of the brackets is called expanding.

Use FOIL.
(x-6)(x-3)-------------------expand
=x^2-3x-6x+18-----------simplify the like terms, -3x and -6x
=x^2-9x+18----------------the answer

(x+3)(x-1)------------------expand
=x^2-x+3x-3--------------simplify the like terms, -x and 3x
=x^2+2x-3-----------------the answer

A. 7(n+4) +6(n+7) So first situations 7 through n. and then 7 through four. the 1st bracket cones out at 7n+28 Do the comparable for the 2d bracket. 6*n =6n 6*7=40 two the two consequences for each brackets are 7n+28+6n+40 two Then simplify the equation. (7n+6n) + (28+40 two) = 13n +70 Then do the comparable approach for the different questions. B. 8(n+6) +4(n-3) 8*n =8n 8*6=40 8 4*n=4n 4*-3=-12 8n+40 8+4n-12 Simplify the equation. =12n-36 C. 5(n-9) -(9-n) 5n-40 5-9+n =6n-fifty 4 desire this helped

1)
(x-6)(x-3)
= x^2 - 6x - 3x + 18
= x^2 - 9x + 18

2)
(x+3)(x-1)
= x^2 + 3x - x - 3
= x^2 + 2x - 3

Simply you Just Remember F.O.I.L.
First term of each bracket
Outer term of each bracket
Inner term of each bracket
Last term of each bracket
Then combine terms

So your first example starts as (x - 6)(x - 3)

First terms x times x
Outer terms -3x
Inner terms - 6x
Last terms -6 x -3
Written out and arranged x² - 9x + 18

Your second example starts as (x + 3)(x - 1)
First terms x times x
Outer terms -x
Inner terms +3x
Last terms -3
Rewrite and arrange x² + 2x - 3

(x-6)(x-3)
xsq-3x-6x+18
xsq-9x+18
8xsq +18

(x - 6)(x - 3)

= x(x - 3) - 6(x - 3)

= x^2 - 3x - 6x + 18(now simplify -3x -6x = - 9x)

= x^2 - 9x + 18.


(x + 3)(x - 1)

= x(x - 1) + 3(x - 1)

= x^2 - x + 3x - 3(now simplify -x + 3x = +2x)

= x^2 + 2x - 3.

Hope this helps, Twiggy.

First Multiply x with (x-3)
then multiply -6 with (x-3)
answer is x2-9x+18

First Question: (x-6)(x-3)

x^2 -9x +18

2nd Question: (x+3)(x-1)

x^2 +2x -3