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).
@robertveith63839 ай бұрын
@ 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.
@IAMswayingwillow9 ай бұрын
I love the way this was presented it makes it fun and it was something I needed to learn
@Fire_Axus6 ай бұрын
your feelings are irrational
@adw1z9 ай бұрын
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!
@KeplerYubified9 ай бұрын
these videos are so good - i love them - next can you do COMPOUND AREA BUT IT KEEPS GETTING HARDER
@pietergeerkens63249 ай бұрын
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.
@ramzfn35439 ай бұрын
jensen math is the best! I love you dude your awesome:)
@djfghjif9 ай бұрын
im in grade 7 and learnt factoring until level 6. how you explain it is really clear and makes it look easier. appreciate it
@sel55959 ай бұрын
what? my little brother is in grade 7 and they are learning ab patterns 💀💀
@Mathmaniac-vw9ip6 ай бұрын
@@sel5595 bro, we learned that in grade 5 💀
@alexfrosa21639 ай бұрын
This was easy for those who did the Kumon maths program. The beginning of level J is full of such factoring tricks and challenges!
@G3R0George9 ай бұрын
For level 6, could you have factored the trinomial final factors into binomial factors?
@alokmishra42939 ай бұрын
No, they are unfactorise-able
@DriftinVr8 ай бұрын
Got them all! Very good quiz on factoring
@antik88679 ай бұрын
We need more videos like this
@robertveith63839 ай бұрын
@ JensenMath -- For Level 6, you misspelled "Parallel."
@RR-bs9mr9 ай бұрын
me who skipped to the end
@riyabhaduri32069 ай бұрын
Same
@williamraleigh75466 ай бұрын
Personally, I can't edge that quickly, so I like to stroke it throughout the entire video. Then I finish at the end.
@jeremyjay3807 ай бұрын
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.
@Fire_Axus6 ай бұрын
this is easy for you because you have them figured out. also where are the non-polynomials?
@DonDak49 ай бұрын
As a 12 years old , I can’t even tying my shoes
@skyofquacks9 ай бұрын
7:13 "Paralell Parking"
@noahvale26279 ай бұрын
Difference is my favorite formula
@bijipeter14719 ай бұрын
Thank you,sir
@realfrogZ9 ай бұрын
im 4th grade and i can do only level one. pretty good right?🙃
@auztenz9 ай бұрын
Yeah! This shows your interested in maths! Keep on going
@georgestoica97449 ай бұрын
Can probably do all of it , just got 120 out of 120 on further mechanics
@mehmetseyhmus9 ай бұрын
Why did you extend it, factor it directly, it can be done even from the mind.
@ataulhaye58019 ай бұрын
love these videos
@muskyoxes9 ай бұрын
Difficulty: any Canadian team winning the Cup
@nkl78739 ай бұрын
Why was the lvl 10 more difficult than lvl 11?(i mean i solved it both and found lvl 11 easier)
On level 5 you don't need to take the x common we can do it without doing that.
@masterofdeception98029 ай бұрын
maybe limits?
@exodus_20_159 ай бұрын
14:50 Ouch.
@sayedizaanahmad9 ай бұрын
Level 6 could have been factored even more into (x-3) (x+3) (x+3) (x+1) (x-1) (x-1)
@adw1z9 ай бұрын
Not true, easy to see since plugging in x = 1 doesn't give 0
@jupitertank71339 ай бұрын
Level 6 after fully factoring has 2 real solutions, -1 & 3 and the rest four solutions are complex pair of conjugated!!!
@Beckham-pw4ee9 ай бұрын
My max is level 7
@botot09 ай бұрын
so nice
@dhamakedarvlog1149 ай бұрын
I wasn able to solve all except 8 and 11
@zxtremedemon9 ай бұрын
I’m in grade 8 so I think getting to level 6 is good for my grade
@zxtremedemon9 ай бұрын
Also, running a marathon is easier than pleasing my parents (at my age I already did that, just before the time limit.)
@shivx32959 ай бұрын
nah atleast till level 9
@riyabhaduri32069 ай бұрын
Level 9,11,5 is easy
@sennpowerhv69229 ай бұрын
I got level 2 but not 1
@shivx32959 ай бұрын
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
@Morax___9 ай бұрын
The last one shouldn't be there
@manitmaniar6169 ай бұрын
All levels can be done by a ninth grade
@shivx32959 ай бұрын
so true even i could do it i am an 8th grader from india
@Ricardo_S9 ай бұрын
A get to level 10
@Thiefy_9 ай бұрын
…
@rakeshsharma59739 ай бұрын
All of these are child level.
@Sbbooster9 ай бұрын
Agreed I’m 8 and did all of them😊
@kevinchen93899 ай бұрын
.
@jammymlbbgameplay9 ай бұрын
Level 11 easy
@cpsc49 ай бұрын
The level 2 needs some fixing the sum must be -3 and the product must be -28. The answers are -7 and 4
@robertveith63839 ай бұрын
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.