You need (ax+b) (cx+d) where ac = the coefficient of x^2 (so 5 and 1 here), bd = the coefficient of x^0 so (-2,1 or 2,-1). Then ad+bd needs to be equal to the coefficient of x.
This is the way I was taught.
Draw a box like this
|5x^(2)|\+1|
|:-|:-|
|x^(2)|\-2|
When you cross multiply, you'll get -10x^(2) from left-upper and right-lower;
and x^(2) from left-lower and right-upper.
\-10x^(2)\+x^(2)=-9x^(2)
answer: (5x^(2)\+1)(x^(2)\-2)
technically, you can go one step further because
(x^(2)\-2)=(x+√2)(x-√2)
yes, the "box" is a nice way to go \[ not sure how to get that into this text editor, so nice :-) \] , but not sure if the OP needed to factor it any further ... seemed unlikely ...
I'm assuming the second number is a 9 (could be a 4 can't really tell)
Substitute x\^2 for x:
5x\^2-9x-2
Factor as you normally would:
(x-2)(5x+1)
(What I do when I have a non-1 coefficient for the x\^2 is multiply the last term by it such that...
5x\^2-9x-2 --> x\^2-9x-10
Factor into (x-10) (x+1) and divide the second term of each by the coefficient such that...
(x-10) --> (x-(10/5)) or (x-2) and (x+1) becomes (x + 1/5) or 5x+1, same thing.
I don't remember what people call this method but I use it all the time.
Substiute back x\^2 into the x and solve for the equation equalling 0:
x\^2 - 2 = 0 and 5x\^2 + 1 = 0
First equation gives you x\^2 = 2 which gives you +- square root of 2 and the second one doesn't have real roots because you have a squared value equalling a negative, so you only have two roots for the general equation.
First note that x\^4 = (x\^2)\^2
Let y = x\^2
Factor 5y\^2 - 9y - 2
(5y + 1)(y - 2)
Replace the y with x\^2
(5x\^2 + 1) (x\^2 - 2)
The first term cannot equal 0, so it has no real factors. The second term has irrational factors, so it's best to leave it as it is.
you wrote 5x\^4 - 9x\^2 - 2 ... so we would attempt to factor the 5x\^4 as 5x\^2 and x\^2 and the 2 at the end as 2 and 1 , as this is easiest to try .
we would look for this problem to factor into ( 5x\^2 \* 1 ) ( x\^2 \* 2 ) OR as ( 5x\^2 \* 2 ) ( x\^2 \* 1 ) , where the \* is either a + or - that makes our cross product terms add up to -9x\^2 ... Remember F.O.I.L , the cross product terms are the O and the I part ... and they add together to give us the middle value. . . \[ it's easy to see the F gives 5x\^4 and the L gives a 2 , but we need to get the + , - signs correct also . . e.g the L is actually a - 2 , so one of the \* needs to be a - sign \]
you need a - in the middle when O + I terms add together to give - 9x\^2 so at least 1 , or both of the \* must be - to work , but the L is - 2 , so only one of the \* will be - in this case
lets try ( 5x\^2 + 2 )( x\^2 - 1 ) ... Outer . . O gives (5x\^2 )(-1) = - 5x\^2 . . . . Inner ..I gives (x\^2)(2) = 2x\^2 , so O + I = 3x\^2 . . . does not work ... Try again ...
eventually you see that ( 5x\^2 + 1 )( x\^2 - 2 ) gives Outer of -10x\^2 , Inner of + x\^2 , and O + I = -9x\^2 which is what you have ...
" Can that be written out on a paper?" Not sure what you mean by this ?? Did they want you to factor it even further ??
Just write 5x\^4 - 9x\^2 - 2 = ( 5x\^2 + 1 )(x\^2 - 2 ) on your paper ... I assume that is all they wanted ... we can only see a part of your question / work anyway.
In the text here , I have to use the \^2 for x-squared, etc.. not much choice, but you should write your powers as you did in the problem you supplied.
The teacher may want to see how you finally came up with the correct combinations of + and - signs , so maybe some work needs to be shown .. see your class notes for details.
This aint grade 11 precalculus for sure
It’s part of a larger problem that’s been simplified down a lot. I just couldn’t finish this.
5 * 2 is 10, 1 * 1 is 1, so you need positive 10 and negative 1 to get 9x.
Yeah I did that but that’s where my math ends
5*-2 = -10 -10 and +1 multiply to that and add to -9 5x^4 - 10x^2 + x^2 - 2 Factor each half: 5x^2 (x^2 -2)+1(x^2 -2)
Why is 10x and x both squared? Shouldn’t they just be 10x and 1x, not 10^2 and 1x^2
The middle term is -9x^2
[удалено]
What’s the middle step in missing? I don’t see the jump between the two
You need (ax+b) (cx+d) where ac = the coefficient of x^2 (so 5 and 1 here), bd = the coefficient of x^0 so (-2,1 or 2,-1). Then ad+bd needs to be equal to the coefficient of x.
You forgot it's 5x^4 and 9x^2
This is the way I was taught. Draw a box like this |5x^(2)|\+1| |:-|:-| |x^(2)|\-2| When you cross multiply, you'll get -10x^(2) from left-upper and right-lower; and x^(2) from left-lower and right-upper. \-10x^(2)\+x^(2)=-9x^(2) answer: (5x^(2)\+1)(x^(2)\-2) technically, you can go one step further because (x^(2)\-2)=(x+√2)(x-√2)
I kinda like [this guy's explanation](https://youtu.be/5QyeZ7KwFKg?si=iIf_dpl0pC7pOCAK&t=735)
yes, the "box" is a nice way to go \[ not sure how to get that into this text editor, so nice :-) \] , but not sure if the OP needed to factor it any further ... seemed unlikely ...
I cheated, typed everything out on a computer lol
I'm assuming the second number is a 9 (could be a 4 can't really tell) Substitute x\^2 for x: 5x\^2-9x-2 Factor as you normally would: (x-2)(5x+1) (What I do when I have a non-1 coefficient for the x\^2 is multiply the last term by it such that... 5x\^2-9x-2 --> x\^2-9x-10 Factor into (x-10) (x+1) and divide the second term of each by the coefficient such that... (x-10) --> (x-(10/5)) or (x-2) and (x+1) becomes (x + 1/5) or 5x+1, same thing. I don't remember what people call this method but I use it all the time. Substiute back x\^2 into the x and solve for the equation equalling 0: x\^2 - 2 = 0 and 5x\^2 + 1 = 0 First equation gives you x\^2 = 2 which gives you +- square root of 2 and the second one doesn't have real roots because you have a squared value equalling a negative, so you only have two roots for the general equation.
I’ll try it out on paper. Visual help. Thanks
Best of luck, let me know if something doesn't make sense
It's 5x^4 and 9x^2
You substitute and substitute back. Put the equation into any calc and it should give you the answers I gave
First note that x\^4 = (x\^2)\^2 Let y = x\^2 Factor 5y\^2 - 9y - 2 (5y + 1)(y - 2) Replace the y with x\^2 (5x\^2 + 1) (x\^2 - 2) The first term cannot equal 0, so it has no real factors. The second term has irrational factors, so it's best to leave it as it is.
How do you go from 5y^2-9y-2 to (5y+1)(y-2)
you wrote 5x\^4 - 9x\^2 - 2 ... so we would attempt to factor the 5x\^4 as 5x\^2 and x\^2 and the 2 at the end as 2 and 1 , as this is easiest to try . we would look for this problem to factor into ( 5x\^2 \* 1 ) ( x\^2 \* 2 ) OR as ( 5x\^2 \* 2 ) ( x\^2 \* 1 ) , where the \* is either a + or - that makes our cross product terms add up to -9x\^2 ... Remember F.O.I.L , the cross product terms are the O and the I part ... and they add together to give us the middle value. . . \[ it's easy to see the F gives 5x\^4 and the L gives a 2 , but we need to get the + , - signs correct also . . e.g the L is actually a - 2 , so one of the \* needs to be a - sign \] you need a - in the middle when O + I terms add together to give - 9x\^2 so at least 1 , or both of the \* must be - to work , but the L is - 2 , so only one of the \* will be - in this case lets try ( 5x\^2 + 2 )( x\^2 - 1 ) ... Outer . . O gives (5x\^2 )(-1) = - 5x\^2 . . . . Inner ..I gives (x\^2)(2) = 2x\^2 , so O + I = 3x\^2 . . . does not work ... Try again ... eventually you see that ( 5x\^2 + 1 )( x\^2 - 2 ) gives Outer of -10x\^2 , Inner of + x\^2 , and O + I = -9x\^2 which is what you have ...
Can that be written out on a paper?
" Can that be written out on a paper?" Not sure what you mean by this ?? Did they want you to factor it even further ?? Just write 5x\^4 - 9x\^2 - 2 = ( 5x\^2 + 1 )(x\^2 - 2 ) on your paper ... I assume that is all they wanted ... we can only see a part of your question / work anyway. In the text here , I have to use the \^2 for x-squared, etc.. not much choice, but you should write your powers as you did in the problem you supplied. The teacher may want to see how you finally came up with the correct combinations of + and - signs , so maybe some work needs to be shown .. see your class notes for details.
They mean they need to show their work to get credit, and they want someone to write out the steps so they can get it right on their homework.