### Chap 6: Anything Wrong?

Antonio, Benedict and Celine were tasked to factorise the following:
40ax - 10bx - 60ay + 15by
and all three presented different answers.

Antonio. (10x - 15y)(4a - b)
Benedict. (2x - 3y)(20a - 5b)
Celine. 5(3y - 2x)(b - 4a)

Discuss the solution by each student and determine if each is correct.
- if the answer is correct, describe how he/ she arrived at that answer
- if the answer is incomplete, how should he/she continue from there
- if the answer is incorrect, point out what is the likely mistake

1. Antonio's answer is correct. He basically grouped the first and second terms as the first group, and grouped the third and fourth group as the second group. After forming the two groups, he factorises them.

Benedict's answer is also correct. He rearranged the terms in this manner:
40ax-60ay-10bx+15by. He then proceeds to factorise them after grouping the first two terms as the first group and the last two terms as the second group.

Celine's answer is correct. She starts by extracting the highest common factor from the four coefficients. Since the first factorisation results in 5(8ax-2bx-12ay+3by), she proceeds to factorise the terms in the brackets by grouping the first two terms in the bracket as the first group and the last two terms in the brackets as the second group. She would then get this: 5[4a(2x-3y)-b(2x-3y)]. Finally she would arrive at this answer: 5(2x-3y)(4a-b). However, her answer is different from mine, no doubt that they result in 40ax-10bx-60ay+15by. Her answer is actually rearranged: 15by-60ay-10bx+40ax.

As of Celine's answers, you are right to say that she has extracted the highest numerical common factor. However, what you've got is different from hers. So, is her answer still acceptable?

2. Celine's answer is acceptable, as it is rearranged version of the original expression. Antonio and Benedict should actually simplify the answers, by extracting the highest common factor from the coefficients as the first step. Their method should be like Celine's, and they should end up with the simplified answer of: 5(2x-3y)(4a-b)

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4. Both Antonio and Benedict are incorrect, as they did not remember to take out the common factor, 5.
Celine is correct.

5. a=10
b=20
x=30
y=40
40ax - 10bx - 60ay + 15by
=40(10*30)-10(20*30)-60(10*40)+15(20*40)
=-6000

Antonio is correct.
(10x - 15y)(4a - b)
=(10(30) - 15(40))(4(10) - 20)
=(300 - 600)(40-20)
=(-300)(20)
=-6000

Benedict is correct
(2x - 3y)(20a - 5b)
=(2(30) - 3(40)(20(10) - 5(20))
=(60 - 120)(200 - 100)
=(-60)(100)
=-60 * 100
=-6000

Celine is correct
5(3y - 2x)(b - 4a)
=5(3(40) - 2(30))(20 - 4(10))
=5(120-60)(20-40)
=5(60)(-20)
=5*60*-20
=300*-20
=-6000

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7. I think Antonio is correct as 40ax - 10bx - 60ay + 15by can be factorised into 10x(4a-b)-15y(4a+b) which can be simplified into (10x-15y)(4a-b).

I think Benedict is also correct as 40ax - 10bx - 60ay + 15by can also be factorised into 20a(2x-3y)-5b(2x-3y) which can be simplified into (2x - 3y)(20a - 5b).

I think Celine is wro0o0o0o0ng, because I cannot understand what goes through her head when she was factorising and I cannot arrive at the same answer as her.

8. Antonio is correct.

If you factorise 40ax - 10bx - 60ay + 15by, you will get (10x-15y)(4a-b). If you expand the answer, you will get 40ax-10ab-60ay+15by, so it is correct.

Benidict is also correct, but it can be factorised further. If he were to expand his answer, he will get 40ax-10ab-60ay+15by also, so he is correct. However, he can factorise (20a - 5b) further, into 5(4a-b)

Celine is also correct, since her answer when expanded further is 40ax-10ab-60ay+15by, so it is correct :)

Sarah Chan

9. Antonio's answer is correct. If we expand his answer we get 40ax - 10bx - 60ay + 15by.

Benedict's answer is correct. If we expand his answer we get 40ax - 10bx - 60ay + 15by.

Celine's answer is correct. If we expand her answer we get 15by - 60ay - 10bx + 40ax, which is a rearranged version of 40ax - 10bx - 60ay + 15by.

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11. I think Celine is correct.

Antonio is wrong because he did not fully factorise. In his answer, (10x - 15y)(4a - b), the (10x - 15y) part could be further factorised into 5(2x - 3y).

Benedict is also wrong because his answer can also be further factorised. His answer is (2x - 3y)(20a - 5b). Though he factorised the the first part, he applied "5", that was factorised out of the first part, to the second part. Now, the second part must be further factorised.

Celine is correct as she fully factorised both sides.

12. Antonio is correct. As 40ax-10bx-60ay+15by
=10x(4a-b)-15y(4a+b)
=10x(4a-b)+15y(4a-b)
=(10x+15y)(4a-b)

And for Benedict and Celine, they are wrong as their answers are different.

13. Antonio, Benedict and Celine were tasked to factorise the following:
40ax - 10bx - 60ay + 15by
and all three presented different answers.

Antonio. (10x - 15y)(4a - b)
This is incomplete because that he used the correct way to factorise, like:
10x(4a-b)-15y(4a-b)
= (10x-15y)(4a-b)
And it expands into 40ax-10bx-60ay+15by
But there is one thing that he forgot: the 5
Benedict. (2x - 3y)(20a - 5b)
This is correct because that he also uses the correct way to factorise, like:
2x(20a-5b)-3y(20a-5b)
And it expands into 40ax-10bx-60ay+15by

Celine. 5(3y - 2x)(b - 4a)
It is correct because that it also expands into 40ax-10bx-60ay+15by
Like 5x3yxb=15yb
5x2Xxb=10bx

14. My Answer : 20a(2x - 3y) - 5b(2x - 3y) = (20a - 5b) (2x - 3y)

The answer to Antonio's factorisation is : (10x - 15y) (4a - b) = 40ax - 10bx - 60ay + 15by. I feel that Antonio was correct as he divided the numbers with the highest common factor.

The answer to Benedict's factorisation is : (2x - 3y) (20a - 5b) = 40ax - 10bx - 60ay + 15by. I feel that Benedict's answer is incomplete as he could further factorise by dividing the numbers with their highest common factor.

The answer to Celine's factorisation is : 5(3y - 2x) (b - 4a) = (15y - 10x) (b - 4a) = 15by - 60ay - 10bx + 40ax. I feel that Celine's answer is incomplete as Celine simplified it after factorising it, dividing the answer further.

15. Antonio's answer is correct because he factorise both sides by grouping them by the terms. Then, he factorises them.

Benedict's answer is also correct as he aslo did quite similar to what Antonio did. However, he used a smaller common factor at the left side and bigger common factor at the right side.

Celine's answer is also correct as she gets the common factor and multiplies with the rest factorisation statement.

16. 1) Antonio's answer is incomplete. He should simplify (10x-15y) to 5(2x-3y).

2) Benedict's answer is incomplete. He should simplify (20a-5b) to 5(4a-b).

3) Celine is incorrect. 5b(3y+2x)(b-4a)= (15by+10bx)(b-4a)=15b^2y-60aby+10b^2x-40abx which is not the same as the above question.

17. Antonio's answer is correct because:

40ax - 10bx - 60ay + 15by
=10x(4a-b) -15y( 4a-b)
=(4a-b)(10x-15y)

40ax - 10bx - 60ay + 15by
= 20a(2x-3y) -5b(2x-3y)
= (2x - 3y)(20a - 5b)

40ax - 10bx - 60ay + 15by
= -20a(-2x+3y) +5b(-2x+3y)
= (-2x + 3y)(-20a + 5b)
= 5(3y - 2x)(b - 4a) -> 5 does not multiply 3y - 2x, but multiplies (b - 4a). Thus, it should be 5(b-4a)(3y-2x)

18. ANS:40ax-10bx-60ay+15by
=10x(4a-b)-15y(4a-b)
=(10x-15y)(4a-b)

Antonio is CORRECT .He arrived to his answer by GROUPING COMMON TERMS . In this case , he grouped the common term '10x' from "40ax-10bx" and 15y from "-60ay+15by".

Benedict is CORRECT . He arrived to his answer by GROUPING COMMON TERMS but he grouped different common terms than Antonio.In this case , he rearranged it the expression” 40ax-10bx-60ay+15by” to “40ax-60ay-10bx+15by”.Thus, he grouped the common terms , ’a’ from “40ax-60ay” and ‘b’ from “-10bx+15by”.Thus,arriving to the answer , “(2x - 3y)(20a - 5b)”

Celine is CORRECT.When she factorised,she rearranged the expression “40ax-10bx-60ay+15by” to “15by - 60ay - 10bx + 40ax”.Thus, when she factorised she GROUPED COMMON TERMS “y” from “15by - 60ay” and “x” from “- 10bx + 40ax” to get ,”5(3y - 2x)(b - 4a)”