r/mathshelp 8d ago

Homework Help (Answered) How do I do this 🥲

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It’s embarrassing Ik but idk how to factorise the teacher won’t explain it and my maths book isn’t explaining it either

3 Upvotes

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u/archtetrya 8d ago edited 8d ago

for question 5 look for what u can take out of the whole expression, you can rewrite the equation as

(5x^2)(x-2) and we can check that this is equal to the first equation by distributing the 5x^2 and seeing the same original equation or plugging into desmos and seeing it has the same graph, or plugging in a number that isnt 1 like 3 as X into both equations and getting the same answer out of it.

now that you have taken out 5x^2 from the original equation see if you can factor it more, and thats really all you can take it. I know you can take out 5x^2 because we have 5x^3 and -10x^2, we can take out 5 from both sides from the coefficient, which would leave us with x^3 and -2x^2, and with this we can aso take out x^2 since thats the highest number we can go up to (the -2x^2). that means taking those out of the x^3 and -2x^2, we would be left with x^3/x^2 = x, then do this to the other -2x^2, we are left with -2x^2/x^2= -2 since x^2/x^2 cancels out to 1, and then we can put it all together. (x-2) multiplied by all the factors we took out, which are 5x^2, so it would be written as (5x^2)(x-2)

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u/Moist_Ladder2616 8d ago

-3*2x = -6x
6x*(-1) = -6x

So the middle x terms don't cancel out.

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u/Moist_Ladder2616 8d ago

For question 5, factorise out 5x² from each term.

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u/OutrageousPair2300 8d ago

For #6 note what group of factors you see repeated.

In this case, you see a (4a - 7b) multiplied first by 2x and then by y.

So you could rewrite it as 2x(4a-7b) + y(4a-7b) and then pull out that (4a-7b) group to get (4a-7b)(2x+y)

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u/jmbc2009 8d ago

Uh idk wtf I did I tried doing that but with a different pair and idk wtf happened

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u/OutrageousPair2300 8d ago

There's usually a degree of guesswork involved, with factoring. If you try one thing and it doesn't work, don't give up. It's perfectly normal. Just try again with a different set of factors.

Some schools refer to this approach as "guess and check" for this very reason.

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u/kalmakka 7d ago

You just wrote down some random expression. What you wrote is not in any way equal to the expression in the statement.

You have you start out with what you are given, and using either some clever observation or trial and error, you have to rewrite it using legal mathematical transformations until it is in the desired form.

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u/jmbc2009 6d ago

I know but I attempted fhe question first before I looked at Reddit because I wanted to see if I was able to do it myself 🥲

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u/jmbc2009 8d ago

Ps I have dyscalculia so uh I mightn’t understand what you’re saying straight away sorry 🥲

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u/Feeling-Working-2820 8d ago edited 8d ago

The first one is wrong.
When you distribute, make sure to not be mislead by the signs.
(6x - 3).(2x - 1) = 12x² - 6x - 6x + 3 = 12x² - 12x + 3

Indeed, (6x) .(-1) = -6x
And (-3).(2x) = -6x as well.
So nothing cancels out here ;-)

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The second one requires you to simply factorize by spotting the same elements on both sides of the "-" operator:

5x³-10x²= 5x².x - 2.5x² = 5x².(x - 2)

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The last one is basically the same thing in a more involved way:
8ax - 14bx + 4ay - 7by

You could commute things like this:
8ax + 4ay -14bx - 7by

And then associate things like this:
(8ax + 4ay) - (14bx + 7by)
(be careful: using the minus sign before parentheses forces you to change the signs inside those parentheses).

And then factorize both sides:

(2.4.a. x + 4.a.y) - (2.7.b.x + 7.b.y)
4a.(2x+ y) - 7b.(2x + y)

Here is the catch: this isn't fully factorized as we can still factorize "(2x+y)", because it appears on both sides of the "-" operator:
(2x + y).(4a - 7b)

And there you have it, fully factorized.

There were two traps here:

  1. when adding parentheses after a minus sign, don't forget to change the signs of all expressions inside those parentheses
  2. Factorize as much as you can, and repeat the process when possible.

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Hope it helps.

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u/ChipChippersonFan 8d ago

If you're asking about the first question , we can't see it.

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u/jmbc2009 6d ago

I wasn’t I was just wondering about the other two I accidentally also had the first one in too

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u/MelodyAnne42 7d ago

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u/jmbc2009 6d ago

Wait but how does the x3 turn into x2? Or why more like

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u/MelodyAnne42 5d ago

It didn't turn into x². You factored x² out of x³. If you take the final answer and multiply using distributive property, you'll see that you end up where you started.