Can anyone factorise X^2-7x-8?
It will be in the form of (x +or- ?)(X +or- ?)
Answers:
(x+1)(x-8)
(X -8) (X + 1)
(x-8)(x+1) is the correct answer.
Yeah, it will be like this:
(x-8)(x+1)
Will :)
(X -8) (X + 1)
The x squared has no coefficient so (x )(x ).
The signs of the other two factors are different, if both positive or both negative you can't multply them to get -8. They need to add or subtract to get -7 for the middle term, so 1 and -8 are it.
(x+1)(x-8)
http://www.khake.com/page47
No it must be in the form AX^2 +BX + C = 0
yes i can;
x^2-7x-8
=x^2-8x+x-8
=x(x-8)+1(x-8)
=(x-8)(x+1)
= x(x-7)-8
= x(x-7) - x(8/x)
= x [(x-7) - 8/x]
= (x-7) - 8
= x - 7 -8
= x - 15
x^2 - 7x - 8
Split 7x in such a manner that when multiply, the result is 8.
here 7x can be written as (8x - 1x). So our equation becomes:
x^2 - (8x - 1x) - 8;
x^2 - 8x +1x - 8;
x(x - 8) +1(x-8); Take (x-8) common
Hence
(x-8)(x+1)
simple :
sum of factors = -7
product = -8
trial and error
so put -7 = -8 +1
x^2-7x-8 = x^2 + x -8 x -8
= x(x+1)-8(x+1)
=(x+1)(x-8)
x^2 - 7x - 8
= x^2 - 8x + x - 8
= x*(X-8) +1*(X - 8)
= (X+1) * (X - 8)
To factor quadratic equations like this, rewrite the equation in the form (x+ _) (x+_). Then look at the sign of the constant term. If it is positive your factors Will be of the form (x+_)(x+_) or (x-_)(x-_). Since it is negative,your factors will be (x+_)(x-_). The constants when multiplied must equal 8 so they must be either 4and 2, or 8 and 1. In these simpler types of problems you can plug in the numbers and see what works. If you do that you will end up with (x+1)(x-8). Check your work by multiplying. You get (x+1)(x-8)=x^2-8x+x-8=x^2-7x-8.
It is easier than anybody has said so far.
Look at the numbers before each term - they are 1, -7, and -8. They would add to zero if the sign of -7 was somehow reversed. But you can do that easily - set x to -1, and the whole thing becomes zero.
This method can even work for cubic or quartic expressions - look for the small neat value of x which makes all the positive terms cancel out the negative ones.
If the expression is zero when x = -1, then (x + 1) is a factor. So at first you write x^2 - 7x - 8 = (x + 1) (Ax + B) with the idea of solving for A and B, but you can immediately see that A = 1 otherwise the x^2 term will be wrong, and B = -8 otherwise the constant term will be wrong. These values give you the correct x term, because (x + 1) is a correct divisor - if it wasn't, then it wouldn't, but at the start we knew that it was.
Cubic and quartic and even higher expressions fall apart the same way if you start with a correct factor, based on a value which makes the expression zero. You just build up the second factor term by term until it gets to the end.
If its x to the power of something minus x minus a number, then the two brackets will be:
(x + something)(x - something)
You have to see which two numbers multiply together to make -8 (a positive number and a negative number), and then see whether those two numbers will add together to make -7.
eg -2 and +4 multiply to make -8, but added together = +2
-4 and +2 multiply to make -8 but added together = -2
-1 and + 8 multiply to make -8 but added together = +7
BUT -8 and +1 multiply to make -8 and added together = -7!
Therefore: (x -8)(x+1).
RULES:
If it's x to the power of something PLUS x PLUS a number:
(x+?) (x+?)
If it's x to the power of something PLUS x MINUS a number / If it's x to the power of something MINUS x MINUS a number:
(x+?)(x-?)
If it's x to the power of something MINUS x PLUS a number:
(x-?)(x-?)
Hope this helps!
x² - 7x - 8
(x - 8)(x + 1)
- - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Check
The foil method
(x - 8)(x + 1) = x² - 8x + x - 8 = x² - 7x - 8
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Answers:
(x+1)(x-8)
(X -8) (X + 1)
(x-8)(x+1) is the correct answer.
Yeah, it will be like this:
(x-8)(x+1)
Will :)
(X -8) (X + 1)
The x squared has no coefficient so (x )(x ).
The signs of the other two factors are different, if both positive or both negative you can't multply them to get -8. They need to add or subtract to get -7 for the middle term, so 1 and -8 are it.
(x+1)(x-8)
http://www.khake.com/page47
No it must be in the form AX^2 +BX + C = 0
yes i can;
x^2-7x-8
=x^2-8x+x-8
=x(x-8)+1(x-8)
=(x-8)(x+1)
= x(x-7)-8
= x(x-7) - x(8/x)
= x [(x-7) - 8/x]
= (x-7) - 8
= x - 7 -8
= x - 15
x^2 - 7x - 8
Split 7x in such a manner that when multiply, the result is 8.
here 7x can be written as (8x - 1x). So our equation becomes:
x^2 - (8x - 1x) - 8;
x^2 - 8x +1x - 8;
x(x - 8) +1(x-8); Take (x-8) common
Hence
(x-8)(x+1)
simple :
sum of factors = -7
product = -8
trial and error
so put -7 = -8 +1
x^2-7x-8 = x^2 + x -8 x -8
= x(x+1)-8(x+1)
=(x+1)(x-8)
x^2 - 7x - 8
= x^2 - 8x + x - 8
= x*(X-8) +1*(X - 8)
= (X+1) * (X - 8)
To factor quadratic equations like this, rewrite the equation in the form (x+ _) (x+_). Then look at the sign of the constant term. If it is positive your factors Will be of the form (x+_)(x+_) or (x-_)(x-_). Since it is negative,your factors will be (x+_)(x-_). The constants when multiplied must equal 8 so they must be either 4and 2, or 8 and 1. In these simpler types of problems you can plug in the numbers and see what works. If you do that you will end up with (x+1)(x-8). Check your work by multiplying. You get (x+1)(x-8)=x^2-8x+x-8=x^2-7x-8.
It is easier than anybody has said so far.
Look at the numbers before each term - they are 1, -7, and -8. They would add to zero if the sign of -7 was somehow reversed. But you can do that easily - set x to -1, and the whole thing becomes zero.
This method can even work for cubic or quartic expressions - look for the small neat value of x which makes all the positive terms cancel out the negative ones.
If the expression is zero when x = -1, then (x + 1) is a factor. So at first you write x^2 - 7x - 8 = (x + 1) (Ax + B) with the idea of solving for A and B, but you can immediately see that A = 1 otherwise the x^2 term will be wrong, and B = -8 otherwise the constant term will be wrong. These values give you the correct x term, because (x + 1) is a correct divisor - if it wasn't, then it wouldn't, but at the start we knew that it was.
Cubic and quartic and even higher expressions fall apart the same way if you start with a correct factor, based on a value which makes the expression zero. You just build up the second factor term by term until it gets to the end.
If its x to the power of something minus x minus a number, then the two brackets will be:
(x + something)(x - something)
You have to see which two numbers multiply together to make -8 (a positive number and a negative number), and then see whether those two numbers will add together to make -7.
eg -2 and +4 multiply to make -8, but added together = +2
-4 and +2 multiply to make -8 but added together = -2
-1 and + 8 multiply to make -8 but added together = +7
BUT -8 and +1 multiply to make -8 and added together = -7!
Therefore: (x -8)(x+1).
RULES:
If it's x to the power of something PLUS x PLUS a number:
(x+?) (x+?)
If it's x to the power of something PLUS x MINUS a number / If it's x to the power of something MINUS x MINUS a number:
(x+?)(x-?)
If it's x to the power of something MINUS x PLUS a number:
(x-?)(x-?)
Hope this helps!
x² - 7x - 8
(x - 8)(x + 1)
- - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Check
The foil method
(x - 8)(x + 1) = x² - 8x + x - 8 = x² - 7x - 8
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