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- Thread starter Davio
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- #1

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- #2

- 1,752

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a = 2

b = 5

c = 3

From there, just plug and chug.

- #3

- 1,752

- 1

If you want to factor it rather than using the quadratic formula, this is how you would go about it.

You need 2 numbers that give you the product of positive 3. Well you obviously only have 2 choices: 1,3 - Also, if you have a positive constant/third-term, then you need either 2 positive numbers or 2 negative numbers.

Next, pay attention to the sign of your middle term: It is positive, so that means your 2 numbers must be positive.

Lastly, you need two numbers that give you the product of your first time: 1,2

Multiply diagonally and add the two numbers, check if it gives you 5. If not, reverse the number for only one of them and try again. Then multiply downwards to see if you get your first and second terms. From there, write it as a product.

Example: a^2+2ab+b^2

[tex]a \ \ \ b[/tex]

[tex]a \ \ \ b[/tex]

ab+ab=2ab - Middle term checks out!

[tex](a+b)(a+b)[/tex]

[tex]2 \ \ \ 1[/tex]

[tex]1 \ \ \ 3[/tex]

You need 2 numbers that give you the product of positive 3. Well you obviously only have 2 choices: 1,3 - Also, if you have a positive constant/third-term, then you need either 2 positive numbers or 2 negative numbers.

Next, pay attention to the sign of your middle term: It is positive, so that means your 2 numbers must be positive.

Lastly, you need two numbers that give you the product of your first time: 1,2

Multiply diagonally and add the two numbers, check if it gives you 5. If not, reverse the number for only one of them and try again. Then multiply downwards to see if you get your first and second terms. From there, write it as a product.

Example: a^2+2ab+b^2

[tex]a \ \ \ b[/tex]

[tex]a \ \ \ b[/tex]

ab+ab=2ab - Middle term checks out!

[tex](a+b)(a+b)[/tex]

[tex]2 \ \ \ 1[/tex]

[tex]1 \ \ \ 3[/tex]

Last edited:

- #4

cristo

Staff Emeritus

Science Advisor

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- I look at the number in front of the p^2 term.. since it is 2, this implies that the factors must be (2p & m)(p & n) (where & denotes + or - to be decided)
- The last sign is positive, thus the signs in the brackets are either both + or both -
- The middle sign is positive, thus both signs in the brackets are +; we have (2p +m)(p +n)
- Now, you want numbers (n,m) such that mxn=3 and 2n+m=5; the former expression gives us a choice of (3,1) and the latter fixes n=1, m=3

- #5

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Ah I'll try that, that looks like what I'm looking for, *goes off and factorises things*

- I look at the number in front of the p^2 term.. since it is 2, this implies that the factors must be (2p & m)(p & n) (where & denotes + or - to be decided)
- The last sign is positive, thus the signs in the brackets are either both + or both -
- The middle sign is positive, thus both signs in the brackets are +; we have (2p +m)(p +n)
- Now, you want numbers (n,m) such that mxn=3 and 2n+m=5; the former expression gives us a choice of (3,1) and the latter fixes n=1, m=3

- #6

cristo

Staff Emeritus

Science Advisor

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No worries! I've moved to homework, since it fits in here better!

- #7

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Once you have P the factors become (p-root1)(p-root2) or (p-root1)^2 if there is just one.

k

- #8

HallsofIvy

Science Advisor

Homework Helper

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If the the polynomial does NOT have factors with integer coefficients, you can still use the quadratic formula to factor. For example, the quadratic formula gives [itex]-1/2+ i\sqrt{3}/2[/itex] and [itex]-1/2- i\sqrt{3}/2[/itex] as roots of p

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