How to factorise x^2 - x + 16
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How to factorise x^2 - x + 16

[From: ] [author: ] [Date: 11-05-25] [Hit: ]
Please help.-When in doubt, factor by completing the square... y = x^2 - x + 16.......
I have reviewed the book & tried examples, but still find this difficult.
I know the factors are 1, 2, 4, 8, & 16 & understand that to find the numbers to put into brackets eludes me. (x ) (x )

To find the terms in brackets they should add together for one & multiply for the other.

I originally had 3x^2 - 3x + 48 but divided by 3.
I can't help thinking I did something wrong.
Please help.

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When in doubt, factor by "completing the square"
... y = x^2 - x + 16

... x^2 - x = - 16
or x^2 - 2(x)(1/2) = - 16
or x^2 - 2(x)(1/2) + (1/2)^2 = - 16 + (1/2)^2
or (x - 1/2)^2 = - 63/4
or x - 1/2 = ± (1/2) √63 i
or x - 1/2 ± (1/2) √63 i = 0

... y = x^2 - x + 16
or y = (x - 1/2 - (1/2) √63 i) (x - 1/2 + (1/2) √63 i)
or y = (1/4) (2x - 1 - √63 i) (2x - 1 + √63 i)

Verified: http://www.wolframalpha.com/input/?i=x%5…

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I just finished doing factoring, this i how i was taught..
For the first part you write:
3(x^2 -3x +48
you don't divide by 3.

Then I made to X bracket then put 48 on top and -3 on the bottom .
then i try to see what to number multiply to make 48 and what two numbers add up to -3 which are what ever number you come up with the numbers. So that should help you, but i cant give the answer.

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When they resist, use the quadratic formula.

(-b +/- sqrt(b^2-4ac))/2a

where a is the number of x^2s, b is the number of xs and c is the constant.

So...

(1 +/- sqrt(1-4(16))/2)
And the formula tells us in order to find this, we would need to take the square root of negatives. So you won't be able to solve this without using imaginary numbers.

If you do use imaginaries, you can go with:

1+/- sqrt(-63)/2
(1+/- sqrt(63)i)/2

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You can't factorise x^2 - x + 16.

If you take your original equation 3x^2 - 3x + 48:
12
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