Factoring

[ymathsohard]

Hi everyone!! I'm new here, and I have the following burning math question:

I've recently introduced myself to factoring. I've been able to grasp the concept and am able to work out most (beginner level) problems. However, there are some problems in the book "All you need in Maths!" by dutch author Jan van de Craats that I just cannot figure. I'm not able to find any examples of these particular type of problems on Youtube or anywhere else on the web. So, hopefully you guys can help me out. Please find the problems in the provided image below (entire 4.44)

My main stumbling block is that I don't understand the underlying logic of how to do a step-by-step work out of these problems. I've looked at lessons about different factoring methods such as: factoring polynomials by grouping, AC method, factoring by finding the GCF; However none of these methods seem to give me the necessary tools/understanding to work the below problems. I've gone as far as looking at the special cases for factoring, and the below expressions in 4.44(image) do not qualify as perfect square trinomials, or are an expression of the difference of two squares.

Example of where I go wrong:

So I know that you start with factoring out the GCF. Makes sense, but which side of the equation do you eliminate the GCF. That is important in order to continue.

Anyway if anyone could please provide me with the underlying logic on how to approach these problems step-by-step, or maybe point me to some theory that I could read ?

Thank YOU SO MUCH

 

[Cubist]

Let's look at question a. In this case it might help if you (partly) expand the expression that is squared...

2(a+3)^2 + 4(a+3)
= 2(a+3)(a+3) + 4(a+3)

Now can you see a common factor? Can you continue?

 

[Steven G]

What if we use u for a+3 in (a)

So we have 2u^2 + 4u. Can you factor this? If yes, then do so and then put back a+3 for u. Don't forget to check if you can factor further.

 

[HelpingHand]

a. 2(a+3)^2 + 4(a+3)

= 2(a+3) (a+3+2) [ Take 2(a+3) common factor from equation so there are (a+2) from first term and 2 from second term left ]
= 2(a+3)(a+5) [Your answer]

b. (a+3)^2 (b+1) - 2(a+3)(b+1)

= (a+3)(b+1) (a+3-2) [ Take (a+3)(b+1) common factor from given equation so there are (a+3) from first term and -2 from second term left ]
= (a+3)(b+1)(a+1) [Your answer]

Using same method you can solve c & d. If your face any difficulty to understand it you can ask me frankly.

Let me help you in e...

e. -2(a+4)^3 + 6(a+4)^2 (a+2)

Lets re-arrange Equation

= 6(a+4)^2 (a+2) - 2(a+4)^3
= 2(a+4)^2 { 3(a+2) - (a+4) } [ Take 2(a+4)^2 common factor from given equation so there are 3(a+2) from first term and -(a+4) from second term left ]
= 2(a+4)^2 { 3a + 6 - a - 4 } [ Evaluate { 3(a+2) - (a+4) } ]
= 2(a+4)^2 { 2a + 2 }
= 2(a+4)^2 { 2(a+1) } [ Take 2 common from{ 2a + 2 } ]
= 2*2 (a+4)^2 (a+1)
= 4(a+4)^2 (a+1) [Your answer]

 

[Steven G]

a. 2(a+3)^2 + 4(a+3)

= 2(a+3) (a+3+2) [ Take 2(a+3) common factor from equation so there are (a+2) from first term and 2 from second term left ]
= 2(a+3)(a+5) [Your answer]

b. (a+3)^2 (b+1) - 2(a+3)(b+1)

= (a+3)(b+1) (a+3-2) [ Take (a+3)(b+1) common factor from given equation so there are (a+3) from first term and -2 from second term left ]
= (a+3)(b+1)(a+1) [Your answer]

Using same method you can solve c & d. If your face any difficulty to understand it you can ask me frankly.

Let me help you in e...

e. -2(a+4)^3 + 6(a+4)^2 (a+2)

Lets re-arrange Equation

= 6(a+4)^2 (a+2) - 2(a+4)^3
= 2(a+4)^2 { 3(a+2) - (a+4) } [ Take 2(a+4)^2 common factor from given equation so there are 3(a+2) from first term and -(a+4) from second term left ]
= 2(a+4)^2 { 3a + 6 - a - 4 } [ Evaluate { 3(a+2) - (a+4) } ]
= 2(a+4)^2 { 2a + 2 }
= 2(a+4)^2 { 2(a+1) } [ Take 2 common from{ 2a + 2 } ]
= 2*2 (a+4)^2 (a+1)
= 4(a+4)^2 (a+1) [Your answer]

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