I'm so confused with the whole factorising process. Please help explain: How do we take out the highest common factors in the expressions: a)5ac+10ab-25ad b)8m-24mn c)30gh+15fg-5fh

5 a *c + 5 a *2b + 5 a *5d

= 5 a ( c + 2b + 5d )

by distributive property of multiplication

8m-24mn

m is common and 24 = 8 * 3
so
8 m * 1 - 8 m * 3n
= 8m ( 1-3n)

30gh+15fg-5fh
5 again and I see nothing else common to all three terms

5 ( 6gh + 3fg - 1fh )

5 a *c + 5 a *2b - 5 a *5d

= 5 a ( c + 2b - 5d )

Factorising is the process of writing an algebraic expression as a product of its factors. The highest common factor (HCF) is the largest factor that can be divided evenly into each term of the expression.

To factorise an expression and take out the HCF, follow these steps:

a) 5ac + 10ab - 25ad

Step 1: Identify the common factors of all the terms.

In this expression, all the terms have a common factor of 5 and a common factor of a. So, we can take out the HCF of 5a.

Step 2: Divide each term by the HCF.

5ac divided by 5a equals c.
10ab divided by 5a equals 2b.
-25ad divided by 5a equals -5d.

Now, we can rewrite the expression:
5ac + 10ab - 25ad = 5a(c + 2b - 5d)

b) 8m - 24mn

Step 1: Identify the common factors of all the terms.

In this expression, all the terms have a common factor of 8 and a common factor of m. So, we can take out the HCF of 8m.

Step 2: Divide each term by the HCF.

8m divided by 8m equals 1.
24mn divided by 8m equals 3n.

Now, we can rewrite the expression:
8m - 24mn = 8m(1 - 3n)

c) 30gh + 15fg - 5fh

Step 1: Identify the common factors of all the terms.

In this expression, all the terms have a common factor of 5 and a common factor of h. So, we can take out the HCF of 5h.

Step 2: Divide each term by the HCF.

30gh divided by 5h equals 6g.
15fg divided by 5h equals 3g.
-5fh divided by 5h equals -f.

Now, we can rewrite the expression:
30gh + 15fg - 5fh = 5h(6g + 3g - f)

Remember that factorising is about finding the common factors and then dividing each term by them. This simplifies the expression and allows us to see the common structure in the terms.