Question

a
In the given figure AD and Be
cef
BE
are the
AABC and DFIIBE
show that C = 4 AC
6) factorise - 8 639-26) 10 (39-26.
11) aca-2b-c) +26c.
mm 295 - 329
Д A
G
B
D‍​

Answer 1

Answer:

ANSWER

Let us first find the HCF of all the terms of the given polynomial 3a

2

bc+6ab

2

c+9abc

2

by factoring the terms as follows:

3a

2

bc=3×a×a×b×c

6ab

2

c=3×2×a×b×b×c

9abc

2

=3×3×a×b×c×c

Therefore, HCF=3×a×b×c=3abc

Now, we factor out the HCF from each term of the polynomial 3a

2

bc+6ab

2

c+9abc

2

as shown below:

3a

2

bc+6ab

2

c+9abc

2

=3abc(a+2b+3c)

Hence, 3a

2

bc+6ab

2

c+9abc

2

=3abc(a+2b+3c).

Step-by-step explanation:

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