Question

Factorizing by Quadratics

Answer 1

\[x ^{2}-26x+165\]

Answer 2

Look for a factor pair of +165 that sums to -26.

Since you want positive constant and a negative coeff on the middle term, you know that you will need both pairs in the factor to be.... ? (both positive? or both negative?)

Once you find the right factor pair, a & b, just set up the binomial factors:

\(\Large (x\pm a)(x\pm b)\)

Just figure out the a, b, and the appropriate signs.

Answer 3

So i will be.. \[\left( x-15 \right) \left( x-11 \right)\] If I am correct?

Answer 4

Yes, very good! :)

Answer 5

Thanks very much :D

Answer 6

Ok this problem is \[x ^{2}-x-20\]

Will this one be a bit different?

Answer 7

Well, same process. The numbers will be different, of course, lol. Remember, just check factoring with multiplication... ALWAYS.

Answer 8

\[\left( x-19 \right) \left( x-1 \right)\]

Answer 9

Welllllllll........ let's check that factoring, by multiplying it back out with FOIL:

\(\Large \left( x-19 \right) \left( x-1 \right)=x^2-x-19x+19=x^2-20x+19\)

Definitely NOT what you started with.

Answer 10

You need FACTOR PAIRS of -20.... remember? What number pairs have a product of 20? (don't worry about the signs, you know they have to be OPPOSITE so that you get a -20 as the last term...)

20=1 * 20 = 2*10 = 4*5

Now which of those, when you use OPPOSITE signs, will SUM to the coefficient of the middle term, which is -1?

Answer 11

Ohh :)
\[\left( x-1 \right) \left( x+20 \right)\]

Answer 12

did you CHECK BY MULTIPLYING? (I know you didn't... you would have seen:

\[\Large \left( x-1 \right) \left( x+20 \right)=x^2 + 19x -20\]

Answer 13

READ ABOVE. Which factor pair of 20 (I listed them for you!) will SUM to -1, with the correctly placed opposite signs?

Answer 14

lol :/ *Facepalms* It was so obvious lol.
\[\left( x-5 \right) \left( x+4 \right)\]

Answer 15

Haha.... yeah, well, we all have those days. YAY, that's it, you got it!

Answer 16

Thanks Very Much :)
I became a fan of yours :D

Answer 17

Great, thanks, and you're welcome :) Happy to help.

Answer 18

There is one more
\[y ^{2} +3y-108\]

Answer 19

I think it will be
\[\left( x-9 \right) \left( x+12 \right)\]

Answer 20

Good job, that's it!

Answer 21

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