FACTORING but it keeps getting HARDER!!! 🧠 (how far can you get???)
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- Опубликовано: 12 июл 2024
- Any math student who hopes to be successful NEEDS to be good at factoring. Test yourself with this video and maybe it will help you improve as well!
Prepare to sharpen your factoring skills as we embark on a journey through 11 math expressions, each one presenting a unique factoring challenge. Whether you're a novice or a seasoned mathematician, this video will put your factoring prowess to the test!
We'll kick things off with familiar territory, exploring common factoring techniques and quadratic trinomial factoring to tackle straightforward expressions. But don't get too comfortable - as we progress, we'll dive into increasingly complex scenarios involving factoring by grouping, sums and differences of cubes, difference of squares, perfect square trinomials, and even perfect cube binomials.
Join me as we unravel each expression, peeling back the layers to reveal its factored form. But here's the twist - I encourage you to pause the video and attempt each factoring challenge on your own before I unveil the solution.
Leave a comment below sharing your progress and which expressions you found particularly challenging. Your feedback is invaluable and helps shape future content! Don't forget to visit my website at jensenmath.ca for additional resources and to stay updated on upcoming videos.
And if you find this video helpful or enjoyable, be sure to like, share, and subscribe for more math tutorials and challenges. Let's continue exploring the fascinating world of mathematics together!
Time stamps:
0:00 level 1
1:00 level 2
1:52 level 3
3:55 level 4
5:26 level 5
7:10 level 6
9:04 level 7
11:34 level 8
14:49 level 9
16:39 level 10
19:00 level 11
#factoring #maths #jensenmath #math
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@ JensenMath -- An alternative beginning to Level 7 is to group them as:
3x^5 - 24x^2 + x^4 - 8x - 4x^3 + 32 =
3x^2(x^3 - 8) + x(x^3 - 8) - 4(x^3 - 8) =
(x^3 - 8)(3x^2 + x - 4) =
etc.
I love the way this was presented it makes it fun and it was something I needed to learn
your feelings are irrational
Factoring expressions involving radicals, as seen in evaluating derivatives of radicals using the limit definition, should be on this list.
The last one is close to x^4+4, which shows that the sum of squares can sometimes be factored (by completing the square on the middle term).
Doing before seeing the answers:
1. 12x^2 (2 + 3x^3)
2. (x+7)(x-4)
3. 2(2x^2 + 3x - 20)
= 2(2x^2 +8x - 5x -20)
= 2(x+4)(2x-5)
4. (2x^2-3y)(x^2-9y)
5. y^2 (x^4 - 3x^3) - z^4 (9x^2 - 27x)
= y^2 x^3 (x-3) - 9z^4 x (x - 3)
= x ( (yx)^2 - 9z^4) (x-3)
= x(xy + 3z^2 )(xy - 3z^2)(x-3)
6. (x^3 - 27)(x^3 + 1)
= (x-3)(x+3)(x^2 + 3x + 9)(x^2 - x + 1)
7. Notice (1,3,-4) and (-8,-24,32):
= x^3 (3x^2 + x - 4) - 8(3x^2 +x-4)
= (x-2)(x^2 + 2x + 4)(3x^2 +x-4)
= (x-2)(x-1)(3x+4)(x^2 + 2x + 4)
8. plugging in -1 gives 0, so equating coeffs:
= (x+1)(3x^3 + bx^2 + cx -12)
b + 3 = 2, b = -1 ; c - 12 = -32 , c = -20
= (x+1)(3x^3 - x^2 - 20x -12)
Plugging in -2 makes second factor 0;
= (x+1)(x+2)(3x^2 + dx - 6)
2d - 6 = -20, d = -7
= (x+1)(x+2)(3x^2 -7x - 6)
= (x+1)(x+2)(x-3)(3x+2)
9. (2x+3y)^2 - (3m + 4n)^2
= (2x+3y+3m+4n)(2x+3y-3m-4n)
10. (2x+3)^3 - 64y^3
= (2x+3-4y)(4x^2 + 12x + 9 + 8xy + 12y + 16y^2)
Can only complete the square on second factor, not simplify further?
11. x^4 + 4x^2 + 16
If u = x^2, then u^2 + 4u + 16 has no real roots. Therefore factorising will most likely take the form:
(x^2 + ax + b)(x^2 + cx + d), where a,c non-zero
x^3: a + c = 0 so a = -c ; now form is: (x^2 + ax + b)(x^2 -ax + d);
x: -ab + ad = 0 so a(d-b) = 0 so d = b
==> form is (x^2 + ax + b)(x^2 -ax + b)
x^2 : 4 = -a^2 + 2b
Have b^2 = 16 so b = +-4 and since a^2 = 2b - 4 > 0, b = 4 and so a = +-2
==> x^4 + 4x^2 + 16 = (x^2 + 2x + 4)(x^2 - 2x + 4)
Edit: I missed the trick in the last part:
(x^2 +4)^2 - (2x)^2 and difference of 2 squares. I should’ve seen this!
Only 4 likes and 0 replies?, I also solved the first 10
jensen math is the best! I love you dude your awesome:)
Thank you. Very nice problems for a morning "factoring kata".
If one knows the Sophie Germain identity (for #11) then IMHO #7 & #8 are the most challenging (or, alternatively, the most work) as 9 and 10 get simpler again.
these videos are so good - i love them - next can you do COMPOUND AREA BUT IT KEEPS GETTING HARDER
Got them all! Very good quiz on factoring
We need more videos like this
Thank you,sir
me who skipped to the end
Same
Personally, I can't edge that quickly, so I like to stroke it throughout the entire video. Then I finish at the end.
love these videos
This was easy for those who did the Kumon maths program. The beginning of level J is full of such factoring tricks and challenges!
im in grade 7 and learnt factoring until level 6. how you explain it is really clear and makes it look easier. appreciate it
what? my little brother is in grade 7 and they are learning ab patterns 💀💀
@@sel5595 bro, we learned that in grade 5 💀
Difference is my favorite formula
@ JensenMath -- For Level 6, you misspelled "Parallel."
For level 6, could you have factored the trinomial final factors into binomial factors?
No, they are unfactorise-able
this is easy for you because you have them figured out. also where are the non-polynomials?
so nice
The last one should be fully factored as:
[(x+2)(x+2)-2x][(x-2)(x-2)+2x]
And yes, I did use his logic.
Can probably do all of it , just got 120 out of 120 on further mechanics
7:13 "Paralell Parking"
I got all the levels except level 8 🔥🔥🔥
Level 11 I can factor it into complex (x²+2-2i√3)(x²+2+2i√3)
As a 12 years old , I can’t even tying my shoes
Difficulty: any Canadian team winning the Cup
me who lost all of them
On level 5 you don't need to take the x common we can do it without doing that.
its the good stuff when its get harder in level 5
14:50 Ouch.
I'd say level 9 is easier than level 8 tbh
Why did you extend it, factor it directly, it can be done even from the mind.
My max is level 7
Level 11 it's actually level 1
x⁴ + 4x² + 16
= (x²)² + (4)(x²) + (4)²
= (x²)² + (2)(4)(x²) + (4)² - (2x)²
= (x² + 4)² - (2x)²
= (x² + 2x + 4)(x² - 2x + 4)
Why was the lvl 10 more difficult than lvl 11?(i mean i solved it both and found lvl 11 easier)
maybe limits?
I wasn able to solve all except 8 and 11
im 4th grade and i can do only level one. pretty good right?🙃
Yeah! This shows your interested in maths! Keep on going
Level 6 could have been factored even more into (x-3) (x+3) (x+3) (x+1) (x-1) (x-1)
Not true, easy to see since plugging in x = 1 doesn't give 0
Level 6 after fully factoring has 2 real solutions, -1 & 3 and the rest four solutions are complex pair of conjugated!!!
I’m in grade 8 so I think getting to level 6 is good for my grade
Also, running a marathon is easier than pleasing my parents (at my age I already did that, just before the time limit.)
nah atleast till level 9
I got level 2 but not 1
Level 9,11,5 is easy
all of them were pretty easy not that tough to factor out i think even my 15 year old fiends could do them
need some more difficult factoring questions
A get to level 10
The last one shouldn't be there
All levels can be done by a ninth grade
so true even i could do it i am an 8th grader from india
…
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Level 11 easy
The level 2 needs some fixing the sum must be -3 and the product must be -28. The answers are -7 and 4
Level 2 does not need any fixing. The middle term is correctly given as positive 3x. Wrong. There are no "answers." There is no equation to be solved. There is just an expression to be factored.
All of these are child level.
Agreed I’m 8 and did all of them😊