I would appreciate verification on the following: 1) Completely factor the expression: a^2 + 4b -ab - 4a (I have (a+b)(a-4) ) 2) completely factor r^2-2r+1 (I have r(r-2)+1 ) Thanks!

2. its easy to check these
just expand the brackets

$(a+b)(a-4) = a^2 - 4a + ba -4b$

which is not your original equation

the second one should be $(r-1)^2$

3. Originally Posted by jay1
Completely factor the expression: a^2 + 4b -ab - 4a
$a^2 + 4b -ab - 4a$

Grouping

$a^2 -ab -4a+ 4b$

$(a^2 -ab) -(4a- 4b)$

Find a common factor in each group

$a(a -b) -4(a- b)$

Can you see the common factor?

4. ## keep in mind

an equation in the form ax^2+bx+c=0 is called a quadratic equation. it can always be written as product of two factors.

1) a^2+4b-ab-4a
=a^2-ab+4b-4a
=a(a-b)+4(b-a)
=(a-b)(a-4) [there are two factors]
2)r^2-2r+1
=r^2-r-r+1
=r(r-1)-(r-1)
=(r-1)(r-1) (again two factors though equal)
=(r-1)^2
so before writing r(r-2)+1 remember quadratic equations can be written as product of two factors