<PRE><B>Question 27662</B>
Perform the indicated divisions.

x^2 - 2x - 35
_____________
    x + 5

I need help on this.

Thank you for your time.

Perform the indicated divisions.

(x^2 - 2x - 35)/(x+5)
=[(x-7)(x+5)]/(x+5) = (x-7)  [cancelling (x+5) in the numerator and in the denominator]
Note: (x^2 - 2x - 35) is a quadratic expression in  x and hence factorising 
and expressing it as a product of two  linear factors in x we get
(x^2 - 2x - 35)= x^2 +( -7x+5x)- 35 = x(x-7)+5(x- 7)= (x-7)(x+5)

HOW?
You must multiply the coefficient of the square term and the constant term along with the signs, 
take the numerical product, find its factors, group them into two sets in such a way  
that their sum  is the mid term coefficient (midterm is the term in x)(while grouping the factors to form the  two required sets care should be taken  to i nclude every factor, that is, you should  not omit a single factor). Then 
1)if the sign of the product is +, and the  middle term (the term in x) if  +,
then give  both the sets the sign + and express the midterm as the sum of these two terms.
For example:If the given equation is (x^2 +12x + 35)
The product  of the coefficient of x^2 and the constant term 35 is (1)X35 = 35 = 1X5 X7
You may leave out the 1 as (1X anything )=the same thing
You observe that (7+5) = 12 and hence write the midterm 12x = (7x+5x)
(x^2 +12x + 35)= x^2 +(7x + 5x) + 35
2) if the sign of the product is +, and the  middle term (the term in x) if  minus,
then give  both the sets the sign minus and express the midterm as the sum of these two terms.
For example:If the given equation is (x^2 -12x + 35)
The product  of the coefficient of x^2 and the constant term 35 is (1)X35 = 35 = 1X5 X7
You may leave out the 1 as (1X anything) =the same thing
You observe that (-7)+(-5) = -12 and hence write the midterm -12x = (-7x)+(-5x)=-7x-5x
(x^2 -12x + 35)= x^2 +(-7x -5x) + 35 = x^2 -7x -5x + 35
3) if the sign of the product is minus, and the  middle term (the term in x) if  +,
then give the sign of the midterm(the term in x) to the larger set and the other sign to the smaller 
For example:If the given equation is (x^2 +2x -35)
The product  of the coefficient of x^2 and the constant term (-35) 
is (1)X(-35) = -35 = -(1X5 X7)
You may leave out the 1 as (1X anything) =the same thing
You observe that (7)+(-5) = 7-5 = 2 and hence write the midterm +2x = (+7x)+(-5x) =7x-5x
(x^2 +2x -35)= x^2 +(7x -5x) -35 = x^2 +7x -5x -35
4) if the sign of the product is minus, and the  middle term (the term in x) if  minus,
then give the sign of the midterm(the term in x) to the larger set and the other sign to the smaller 
For example:If the given equation is (x^2 -2x -35)
The product  of the coefficient of x^2 and the constant term (-35) 
is (1)X(-35) = -35 = -(1X5 X7)
You may leave out the 1 as (1X anything) =the same thing
You observe that (-7)+(5) = 5-7 = -2 and hence write the midterm -2x = -7x+5x
(x^2 -2x -35)= x^2 +( -7x +5x) -35 = x^2 -7x +5x -35

What is the next step?
Group the four terms as sum of two sums
And use the rule:ap+bp = p(a+b) where p will be your common sum.
Let  us  see the above  4 examples
1) (x^2 +12x + 35)
= x^2 +(7x + 5x) + 35
= (x^2 +7x) + (5x + 35) (by additive associativity)
=x(x+7) +5(x+7) [pulling out x common in the first two terms  and  +5common in the next two]
=xp+5p where p =(x+7)
=p(x+5)
=(x+7)(x+5)
2) (x^2 -12x + 35)= x^2 +(-7x -5x) + 35
= x^2 -(7x + 5x) + 35
= (x^2 -7x) -5x+ 35      (by additive associativity)
=x(x-7) -5(x-7) [why have we pulled out -5 as common instead of +5? 
It is because our aim is to get our p as (x-7). So there is need for pulling out -5]
= xp-5p where p =(x-7)
= p(x-5)
= (x-7)(x-5)
3) (x^2 +2x -35)= x^2 +(7x -5x) -35 
= (x^2 +7x) -5x -35
=x(x+7)-5(x+7) [pulling out x common in the first two terms  and  -5 common in the next two]
=xp-5p Where p = (x+7)
=p(x-5)
=(x+7)(x-5)
4) (x^2 -2x -35)= x^2 +(5x -7x) -35 
= (x^2 -7x) +5x -35 (by additive associativity)
=x(x-7)+5(x-7) [pulling out x common in the first two terms  and  +5 common in the next two]
=xp+5p Where p = (x-7)
=p(x+5)
=(x-7)(x+5)
How do you verify?
1)(x^2 +12x + 35) = (x+7)(x+5)
Consider our answer (x+7)(x+5)
Apply expansion of the  product of two linear factors
How is expansion done?
Multiply the first term of the first bracket(along with the sign) with the whole of the next bracket and again similar treatment to the second term of the first bracket. You  get four terms of which the second and the    third are like terms. They are the terms in x. Add them up. 
The final answer is the sum of a term in x^2, a term in x and the constant term 
which is the quadratic expression given.
1)x^2 +12x + 35=(x+7)(x+5) 
RHS = (x+7)(x+5) = x(x+5)+7(x+5)= x^2 +5x +7x +(7X5)= x^2 +12x + 35 = LHS
2)(x^2 -12x + 35) = (x-7)(x-5)
RHS =(x-7)(x-5) = x(x-5)-7(x-5)= x^2 -5x -7x +(-7)X(-5)= x^2 -12x + 35= LHS
3)(x^2 +2x -35)=(x+7)(x-5)
RHS =(x+7)(x-5) = x(x-5)+7(x-5)= x^2 -5x +7x +7X(-5)= x^2 +2x -35= LHS
4)(x^2 -2x -35)=(x-7)(x+5)
RHS =(x-7)(x+5)= x(x+5)-7(x+5) = x^2 +5x -7x +(-7)X5= x^2 -2x -35= LHS</PRE>