Factorise expression HELP: Does 8ax + 40cx + 3a +18c equal (8x + 3)(a + c) ?

englewood

New member
Hello,

I hope you can help me, I'm trying to Fully Factorise this Expression.

8ax + 40cx + 3a +18c

My result is below

8ax + 40cx + 3a + 18c =
8ax + 3a + 40cx + 18c =

a (8x + 3) + c (40x + 18)=
(8x+3) (a+c)

Is this correct, finding this abit hard.

Last edited:

tkhunny

Moderator
Staff member
Hello,

I hope you can help me, I'm trying got Fully Factorise this Expression.

8ax + 40cx + 3a +18c

My result is below

8ax + 40cx + 3a + 18c(8ax + 3a) + (40cx +18c)a (8x + 3) + c (40x + 18)(8x+3) (a+c)

Is this correct, fining this abit hard.
What rules are you using? Can you UNfactorize and get back to where you started?

Try grouping things that are alike.

8ax + 40cx + 3a + 18c

8ax + 3a + 40cx + 18c -- What rule did I use to switch the two middle terms?

a(8x + 3) + 40cx + 18c -- What rule did I use to separate the 'a' from the two terms in the parentheses?

a(8x + 3) + c(40x + 18) -- Again, what rule?

Don't guess. Use rules.

Jomo

Elite Member
Hello,

I hope you can help me, I'm trying got Fully Factorise this Expression.

8ax + 40cx + 3a +18c

My result is below

8ax + 40cx + 3a + 18c(8ax + 3a) + (40cx +18c)a (8x + 3) + c (40x + 18)(8x+3) (a+c)

Is this correct, fining this abit hard.
Of course it is hard without putting equal signs. How can you tell where an expression ends and another begins?

8ax + 40cx + 3a + 18c
= (8ax + 3a) + (40cx +18c)
= a (8x + 3) + c (40x + 18) = (8x+3) (a+c)

Since
(40x + 18) is not (8x+3) you can't factor out (8x+3) from c (40x + 18).

Maybe try to group differently

Last edited:

Dr.Peterson

Elite Member
Hello,

I hope you can help me, I'm trying got Fully Factorise this Expression.

8ax + 40cx + 3a +18c

My result is below

8ax + 40cx + 3a + 18c = (8ax + 3a) + (40cx +18c) = a (8x + 3) + c (40x + 18) = (8x+3) (a+c)

Is this correct, fining this abit hard.
Note that your final result expands to 8ax + 8cx + 3a + 3c, so it can't be correct.

Are you sure you copied the expression correctly? It could be factored if it were changed to 8ax + 40cx + 3a + 15c.

englewood

New member

Ive grouped the terms with A in them then with C in them.

8ax + 40cx + 3a + 18c
8ax + 3a + 40cx + 18c (grouped terms with a and C)

Ive then factored out the A and C.

a(8x+3) + c(40x+18) (factored out A and C).

I cant factor anymore as there is no common factor in 40 and 18, so that must be the answer? a(8x+3)+c(40x+18)

Ive been told the Expression is correct and there is no mistakes, i assumed the same thing!.

Deleted

Last edited:

Subhotosh Khan

Super Moderator
Staff member
Ive grouped the terms with A in them then with C in them.
8ax + 40cx + 3a + 18c
8ax + 3a + 40cx + 18c (grouped terms with a and C)

Ive then factored out the A and C.
a(8x+3) + c(40x+18) (factored out A and C).
I cant factor anymore as there is no common factor in 40 and 18, so that must be the answer?........................... a(8x+3)+2c(20x+9)
Ive been told the Expression is correct and there is no mistakes, i assumed the same thing!.
Fully Factorise this Expression....................8ax + 40cx + 3a +18c
As far as I can tell the expression above cannot be factorized fully

englewood

New member
As far as I can tell the expression above cannot be factorized fully

englewood

New member
BAD PRACTICE to update/edit an original post after
everything gets confused...

Sorry, I'm new to forums.

Staff member

englewood

New member
You didn't update with the answers to my questions.

8ax + 3a + 40cx + 18c -- What rule did I use to switch the two middle terms? (grouping like terms rule)
a(8x + 3) + 40cx + 18c -- What rule did I use to separate the 'a' from the two terms in the parentheses? a/a
a(8x + 3) + c(40x + 18) -- Again, what rule?
c/c

Last edited:

JeffM

Elite Member
8ax + 3a + 40cx + 18c -- What rule did I use to switch the two middle terms? (grouping like terms rule)
a(8x + 3) + 40cx + 18c -- What rule did I use to separate the 'a' from the two terms in the parentheses? a/a
a(8x + 3) + c(40x + 18) -- Again, what rule?
c/c
Your purpose was to group terms with a common factor, but the name of the rule that you used is the commutative law of addition, meaning

$$\displaystyle j + k = k + j.$$

The rule you used in the other two steps was the distributive rule of multiplication over addition, meaning

$$\displaystyle km + kn = k(m + n).$$

I am in the minority by believing that memorizing the names for these rules is far less important than knowing the rules and how to use them.

Here are some of the most basic rules.

$$\displaystyle a + b = b + a.$$ Additive commutativity.

$$\displaystyle a * b = b * a.$$ Multiplicative commutativity.

$$\displaystyle a + (-\ a) = 0.$$ Additive inverses.

$$\displaystyle a + 0 = a \implies a + (b - b) = a.$$ Additive identity.

$$\displaystyle a * \dfrac{1}{a} = 1 \implies \dfrac{a}{a} = 1.$$ Multiplicative inverses.

$$\displaystyle a * 1 = a \implies a * \dfrac{b}{b} = a.$$ Multiplicative identity.

$$\displaystyle a * 0 = 0.$$

$$\displaystyle ab + ac = a(b + c).$$ Distributivity.

$$\displaystyle ab = 0 \implies a = 0 \text { or } b = 0 \text { or } a = 0 = b.$$ Zero product property.

There are other rules of numbers, but a lot of algebra is just using the rules above to put expressions into a more useful form.

englewood

New member
Thank you for your help and information provided.