Best way to solve such questions is determine one root then use synthetic or long division to get rest of them Here -1/3 is a clear root as 3(-1/3)^3+(-1/3)^2-15(-1/3)-5 = 0 And then do synthetic division, Of main eq with (x+1/3) factor You get 3x^2-15=0 X = +√5, -√5, -1/3
There are lots of ways to do these. You can use the factor theorem and then polynomial long division, you can use the sum/product formulae for polynomials (sum of the roots of a cubic equal -b/a, product of the roots equal -d/a etc) and sometimes you can even use trigonometry. The cubic formula is only really used to completely brute force it if nothing else is working.
That makes no sense at all. I have no idea what your teachers are teaching you. A difference of two perfect squares almost screams to be simplified more. It leaves the final result in its most simplest form.
Oh easy factory by first doing X^2(3x+1)-5(3x+1) (X^2-5)(3x+1) You could factor more and could make it (sqrt(5)^2) and be a different of squares (X+sqrt(5))(x-sqrt(5))(3x+1) But also this is a very lucky problem with having just 3x+1 since if you had a different set then it becomes way longer to factor
And be not conformed to this world: but be ye transformed by the renewing of your mind, that ye may prove what is that good, and acceptable, and perfect, will of God. Romans 12:2 KJV Jesus saves..
There also is a cubic formula. As an example: x^3 - 9*x + 28 = 0 D = -9^3/27 + 28^2/4 = 169 x = cbrt(-28/2 + sqrt(169)) + cbrt(-28/2 - sqrt(169)) x = cbrt(-1) + cbrt(-27) x = -4
(X-3)(X-6) Multiply it out and you get X^2-9X+18 There is a nice method called the ac method from some textbooks which motivates you to factor by grouping. ax^2+bx+c is a general form of 2nd degree polynomials where a,b, and c are real numbers, otherwise known as coefficients. Look at the factors of a and c multiplied together Which is 1*18=18 where a=1 and c =18 From here, choose from the factors of a*c=18 and find the factors that add up to the middle coefficient b=-9 The factors to choose are -3 and -6 because (-6)*(-3)=18 and (-6)+(-3)= -9 From here, you can rewrite x^2-9x+18 as x^2 - 6x - 3x + 18 and can factor by grouping as this video shows. Hope this helps
Why don’t we stop at x^2 - 5? I don’t usually see irrationals in factors; or complex numbers either. But this method does help in identifying the roots without needing the quadratic formula.
No in india this is not called middle term split it is called common method only . Though for cubic polynomials we use middle term splitting but this solution is by common method . In middle term splitting we do something different.
5 = root(5) Multiplied by root(5) So [root(5)]^2 or root(5) squared is equal to 5 Perfect squares are usually considered to be whole numbers, but there are also numbers that are squares and not whole numbers Think about root(2) If you square root(2), you get 2 Or root(2)*root(2)=2 Same goes for any number under the square root
Since a positive sign was taken in front of the the group, hence the signs of the terms remain unchanged. If it would have been a negative sign, then it would have been (3x³ + x²) - (15x + 5).
Not all polynomials can be factored using this method, but when you can see when to use it, it becomes useful. There is something called the factor theorem which utilizes a technique called synthetic division. This is usually taught in algebra 2 courses Some texts introduce it early in algebra 1 but do not tell the whole story. Also wikipedia makes it way more complicated than it should be. But simply put, you can look at the product of the first and last numbers, known as coefficients and use the factors of that product to apply synthetic division to further simplify what you brought into question X^3-6X^2+11X-6 = (X-1)(X-2)(X-3) 1*(-6)= -6 and possible factors of -6 are 1,-1,2,-2,3,-3,6 and -6. The numbers to use synthetic division here are 1, 2 and 3 Once solved, X=1,2,3, which implies that (X-1), (X-2) and (X-3) are factors Therefore (X-1)(X-2)(X-3)=X^3-6X^2+11X-6
Why would you factor with the difference of squares on x^2-5? 5 isn't a perfect square, adding square roots into the polynomial gets confusing and it is pretty useless.
5 is still root(5) Multiplied by root(5) Perfect squares are associated with whole numbers, moreover they are associated with the natural numbers Any real number is still a square of another number Root(pi) is also a real number, and when you square it, it becomes pi But non of these numbers are natural numbers
It was lucky that 3x+1 was a common factor, presumably that was the key.
Yeah otherwise we would have to the long ass method
@@Laptopsigmagamer frrr
@@Laptopsigmagamer what method?
@@shiizu. the remainder theorem
@@tarushiagarwal4276no the dividing thing get one of the real roots then use synthetic division etc.
Best way to solve such questions is determine one root then use synthetic or long division to get rest of them
Here -1/3 is a clear root as
3(-1/3)^3+(-1/3)^2-15(-1/3)-5 = 0
And then do synthetic division,
Of main eq with (x+1/3) factor
You get 3x^2-15=0
X = +√5, -√5, -1/3
Watching from India 🇮🇳
Watching from south africa 🇿🇦
Watching from Pakistan
This guy teaches better than our school teachers @@AlphaSigmaGaming
Me from india
@@ZxDON.Real😢
Your handwriting is good
I doubt if that method applies to every third-degree polynomial.
It doesn't.
Only if the ratio between the first degree and the second degree is the same as the ratio between the third degree and the fourth degree.
There's a cubic formula but... Let's NOT get into it, you'll have nightmares.
It doesn't
Sometimes you need to do more than 5 grouping just to find few solutions
There are lots of ways to do these. You can use the factor theorem and then polynomial long division, you can use the sum/product formulae for polynomials (sum of the roots of a cubic equal -b/a, product of the roots equal -d/a etc) and sometimes you can even use trigonometry. The cubic formula is only really used to completely brute force it if nothing else is working.
I hope these sums come to my Indian University exams 😂😂
Bro😂
No one cares about your university
@@samiatabassum9463 no one cares about your opinion either 😂
@@anmol-e7u okay little minded hobgoblin
@@anmol-e7u yoo bro 💀💀💀
I wish JEE, ISI, IOQM, VITEEE, NEST asks these type of questions.
Bro 😂😂....
EGE
Send it to NTA!
AIR1 IITB CSE😂
Imagine AIR 247 reading this comment 🤣🤣
I just learned this in algebra 2. Except unless it's a difference of two PERFECT squares we don't factor it more
That's what i thought too, but personally works for me bc i realise school could ask for me 3 factors and not just 2 anyways lol 😂 works in that case
Same but I learned in 7th honors algebra
That makes no sense at all. I have no idea what your teachers are teaching you.
A difference of two perfect squares almost screams to be simplified more.
It leaves the final result in its most simplest form.
Then please solve this by using this method : factories: a^3-6a^2+11a-6
Oh easy factory by first doing
X^2(3x+1)-5(3x+1)
(X^2-5)(3x+1)
You could factor more and could make it (sqrt(5)^2) and be a different of squares
(X+sqrt(5))(x-sqrt(5))(3x+1)
But also this is a very lucky problem with having just 3x+1 since if you had a different set then it becomes way longer to factor
If youre gonna factor the x^2-5 you might as well pull a three out of 3x+1 so all the factors in parens are in terms of the roots of the polynomial
😊😊😊😊
😊
💛Nice method 🧡💥💫
Ans:(x²-5)(3x+1)
I was curious how this solved in UA-cam shorts.
Nice handwriting ❤
Thank you bhaiya ❤
Wow what a good explanation 😮
Polynomiuuuhhl. Tree treee treeeeee.
And be not conformed to this world: but be ye transformed by the renewing of your mind, that ye may prove what is that good, and acceptable, and perfect, will of God.
Romans 12:2 KJV
Jesus saves..
Thanks dude 😎
Nice explanation sir🎉
Grear explanation. This is a simple as it can get
watching from Bangladesh 🇧🇩 ❤
Thanks you have made my doubt a huge doubt
Can you also solve it by using Bézout's theorem?
no bro , this method are apply in only few questions but not in all questions lik3 ::-- x^3-3x^2-9x-5 and many more........
How do we do these types please
Nice method dude
Keep it up
Great 👍
All I knew before was combining like terms
Thanks!
Excellent
Very very thanks. I was struggling in this problem but your solution gave me an idea about this. Thank you once again. 😊😊😊😊😊😊❤❤❤❤❤❤
Incredible🔥
sometimes this method is just confusing so better do synthetic division.
Brings me back.
Where did you get the one from?
Thank you🙏💕❤
Thank you ❤
This only works if it nicely factors. Otherwise its algebraic long division and guessing the initial root
There also is a cubic formula.
As an example:
x^3 - 9*x + 28 = 0
D = -9^3/27 + 28^2/4 = 169
x = cbrt(-28/2 + sqrt(169)) + cbrt(-28/2 - sqrt(169))
x = cbrt(-1) + cbrt(-27)
x = -4
@@carultch that's a depressed cubic.
Thank you sir
Bro you made me understand this shit thanks bri
It's an amazing method to solve the problen
Hi , how do you do
x^2-9x+18
(X-3)(X-6)
Multiply it out and you get X^2-9X+18
There is a nice method called the ac method from some textbooks which motivates you to factor by grouping.
ax^2+bx+c is a general form of 2nd degree polynomials where a,b, and c are real numbers, otherwise known as coefficients.
Look at the factors of a and c multiplied together
Which is 1*18=18 where a=1 and c =18
From here, choose from the factors of a*c=18 and find the factors that add up to the middle coefficient b=-9
The factors to choose are -3 and -6 because (-6)*(-3)=18 and (-6)+(-3)= -9
From here, you can rewrite x^2-9x+18 as
x^2 - 6x - 3x + 18 and can factor by grouping as this video shows.
Hope this helps
Thank you sir❤
- FROM INDIA
Thanks a lot ❤
thanks i need this so much thanks
I live in India we learnt this in 5 th grade 😂
Nice 👍
You can already see that this is one of the easier to factor 3rd degree polynomials
thank you
Why don’t we stop at x^2 - 5? I don’t usually see irrationals in factors; or complex numbers either. But this method does help in identifying the roots without needing the quadratic formula.
You are supposed to stop at that. I don’t know why he continued.
When you know what’s 3x+1 really about
Indians🗿
👇
Nice 😊❤
thankyousomuch! I was trying to solve question for hours and your this method just help me out thankyou
Does the square root is applicable? Why can't just write (x+5)(x-5)?
Cuz (x+5)(x-5) = x^2 - 25.
Using a^2 - b^2 = (a+b)(a-b), a = x and b = sqrt(5) since sqrt(5)^2 = 5
X^3-8x^2+4x+48=0
Best way
Factoring by grouping method
Mf taught me more than math teacher in 20 seconds
that's wrong , it's just a designed example to get (x+3) a common factor
You learn math mostly by doing exercises yourself, not from listening to others
Solve this x^3-6x^2+3x+10 and no cheating
In my country INDIA it is middle term split bro .
No in india this is not called middle term split it is called common method only . Though for cubic polynomials we use middle term splitting but this solution is by common method . In middle term splitting we do something different.
Amazing 👍
(3x+1)(x^2-5)=(3x+1)(x-√5)(x+√5)
Meanwhile indians and South Koreans kids be like ..... Why he lifting pen for such an easy question
how to factorize x³-x²-17x-15
(x+1)(x+3)(x-5)
Please help me with matrix 2by 2
I had a mini heart attack when I heard "3x+1"
Thanks sir
Polynomialllll😮
Bro just showed me how to solve my calculus problems in a simple manner without banging my head against the wall, I appreciate it🐧
I am glad to find a very good explanation video 🤩🤩
Thanks!!!!
It all went to shit when they introduced Alphabets to math...
thanks bro
it was very easy
A sta je resenje ovog zadatka?
Is this really the final answer?
Yes.
We are studying it in class 8th of India
Hmn
After 12 hours of work, six hours of derivative exercises in math... I can tell you that my brain is now broken lol
Thx
If it is a factor theorom?
🎉 good sir
Just use factor theorem
It just simple bro 😅 . Solve jee advanced question
I can't understand for your solution
How root 5 came?
5 = root(5) Multiplied by root(5)
So [root(5)]^2 or root(5) squared is equal to 5
Perfect squares are usually considered to be whole numbers, but there are also numbers that are squares and not whole numbers
Think about root(2)
If you square root(2), you get 2
Or root(2)*root(2)=2
Same goes for any number under the square root
not working 🙄
Hey man try to be patient
Maza a gyoo
Where did the 3x∆3 go to
just use foil..
Man everytime I see these, it looks like gibberish. But as he starts working them out, I begin to remember doing this shit in hs lmao
Lol, maths be like- u can't make fun of me
Simple
I had a question the -5 when grouping is not convert into +5 ?
Since a positive sign was taken in front of the the group, hence the signs of the terms remain unchanged. If it would have been a negative sign, then it would have been (3x³ + x²) - (15x + 5).
Don't you need both terms in the binomial to have a square root non-involving decimals for it to be a perfect square?
bro right but what about this x^3-6x^2+11x -6🎉🎉
Not all polynomials can be factored using this method, but when you can see when to use it, it becomes useful.
There is something called the factor theorem which utilizes a technique called synthetic division.
This is usually taught in algebra 2 courses
Some texts introduce it early in algebra 1 but do not tell the whole story.
Also wikipedia makes it way more complicated than it should be.
But simply put, you can look at the product of the first and last numbers, known as coefficients and use the factors of that product to apply synthetic division to further simplify what you brought into question
X^3-6X^2+11X-6 = (X-1)(X-2)(X-3)
1*(-6)= -6 and possible factors of -6 are 1,-1,2,-2,3,-3,6 and -6.
The numbers to use synthetic division here are 1, 2 and 3
Once solved, X=1,2,3, which implies that (X-1), (X-2) and (X-3) are factors
Therefore
(X-1)(X-2)(X-3)=X^3-6X^2+11X-6
thanks
To teach the method, you have to card stack by creating a workable polynomial before you turn the camera on. 😅
Why would you factor with the difference of squares on x^2-5? 5 isn't a perfect square, adding square roots into the polynomial gets confusing and it is pretty useless.
5 is still root(5) Multiplied by root(5)
Perfect squares are associated with whole numbers, moreover they are associated with the natural numbers
Any real number is still a square of another number
Root(pi) is also a real number, and when you square it, it becomes pi
But non of these numbers are natural numbers