Could someone please work this problem for me. It will be greatly appreciated.?

Question

Your friend is taking an algebra class that began on week after yours. Expleain to your friend a general strategy for factoring in a very organized way.

Answer 1

you either take out the common factors or you multiply the extreme terms and divide it into two or more terms that when added up give you the middle term and when multiplied give you the sum of the extreme terms. e.g.
3x^2 + 4x + 1

multiply the extreme terms

3x^2

break it down into
3x and x (multiply them and you get 3x^2, add them and you get 4x)


then replace the original equation's middle term with this

3x^2 + 3x+ x +1

then factorize

3x(x+1) + 1(x+1)

therefore the end result would be

(3x + 1) (x+1)

Answer 2

It depends on what you are trying to factor.

Let us say that you are factoring a quadratic equation of the form

a x^2 + b x + c

1) find two numbers (call them p and q) that multiply to make a
(if a = 1, then p=1=q will often work).

2) find two numbers (call them r and s) that multiply to make c.

3) Then try to combine (p,q) and (r,s) such that p*r + q*s = b
Try all combinations. First product must have one from (p,q) and one from (r,s), the second must be the remaining ones.

4) Once you have identified the 4 numbers, then you have either:
p*r + q*s = b
in which case the factors are (p x + s)(q x + r)
or
p*s + q*r = b
in which case the factors are (p x + r)(q x + s)

5) Verify (always verify)
Let's use the last one in 4:
(p x + r)(q x + s) =
(pq)x^2 + (ps) x + (qr) x + rs

we know, from our work above:
pq = a
rs = c

a x^2 + (ps) x + (qr) x + c

a x^2 + ( (ps)+(qr) )x + c

We know that ps + qr = b

a x^2 + b x + c

Done

Answer 3

Look for clues to break it down. First, most obvious, if it's even, one of its factors is two.

If the sum of the digits is divisible by three, one of the factors is three.

If the sum of the digits is divisible by nine, one of the factors is nine.

If it ends in zero or five, one of the factors is five.

Add the digits in the odd places--the first, third, fifth, and so forth--and then add the digits in the even places. If the two sums are equal or if their difference is divisible by eleven, the number is divisible by eleven.

The largest factor you need to look for cannot be larger than the square root of the number.

Apply each of these, dividing the number by the first factor you find, then dividing the result by the next one that you find, and you'll eventually find all of the factors.

Answer 4

first we look for a common
factor in the expression:
2x^2-4x+2=2(x^2-2x+1)
further factorising:
2(x^2-x-x+1)
=2(x(x-1)-1(x-1))
=2(x-1)^2
we have these basic
formulae to factorise:
(a-b)^2=a^2 -2ab+b^2
(a+b)^2=a^2+2ab+b^2
a^2-b^2=(a+b)(a-b)

Answer 5

I WILL TELL HIM TO BE PERFECT WITH THE IDENTITIES
AND DO PRACTISE
THAT WILL DO