Question

How do you factor completely?: -t^2-9t+1

Answer 1

you cannot factor it in integers, i think we call such polynomials as prime polynomials

Answer 2

so, let us use real numbers :)

Answer 3

PRIME was not a correct response

Answer 4

what have you tried @v_nessa_87 ?

Answer 5

Factoring out -1, trying to apply perfect square trinomials method but the -9t is throwing me off.

Answer 6

so, let's see: if I said this is of the form \(ax^2+bx+c\) what immediately jumps to mind?

Answer 7

FOIL

Answer 8

what else?

Answer 9

That's it. I tried rewriting it like this: -1(t2+(3^2)t-(1^2)) but still don't think that's the right way to go.

Answer 10

just wait, you're ignoring the obvious

Answer 11

if you had \(ax^2+bx+c=0\) how would you get the x values?

Answer 12

Nothing multiplies to -1 and adds to +9.

Answer 13

well I think you need to use the discriminant to investigate if it can be factored
the discriminant is
\[\Delta = b^2 - 4ac\]

Answer 14

That doesn't matter

Answer 15

There is a formula, you probably have memorized

Answer 16

usually starts with a Q

Answer 17

I don't want the value of T, I want to factor.

Answer 18

Isn't the q form for values?

Answer 19

Factoring and finding the value go hand in hand

Answer 20

if \[\Delta > 0\] and is a square number then the quadratic has 2 unequal roots, and it can be factored

Answer 21

and I do not know a Q form for values. No, like seriously, in 6/8th gradish how would you have found the values of x in \(ax^2+bx+c=0\) What formula would you have used?

Answer 22

so use the discriminant so you don't waste you time trying to factor

Answer 23

quadratic formula. i abbreviated: q.uadratic form.ula

Answer 24

YEssss

Answer 25

so, apply it here, once you do

Answer 26

you get two values, those are your second terms in your factoring

Answer 27

I recommend pulling out the -1 first though

Answer 28

you will get a real answer

Answer 29

so given you are asked to factor... the quadratic is prime since \[\Delta = (-9)^2 - 4 \times (-1) \times 1 = 85\] not a square number

Answer 30

You do not need an integer solution to factor... I don't know why you are implying that

Answer 31

-9/2 & sqrt(85)/2

Answer 32

-(t-9/2)(t+sqrt(85/2) ?

Answer 33

actually no, I don't buy it

Answer 34

you shouldn't have a 9 for one and a \(\sqrt{85}\) for the other

Answer 35

check your quadratic answers, but you set it up well

Answer 36

The only reason I am hesitant using the quadratic formula is generally assignments don't have you use skills not yet taught in previous lessons. The quadratic formula isn't introduced for another two lessons ahead of this assignment.

Answer 37

well, I will say it is how I'd do this. Otherwise they want you to complete the square

Answer 38

(I don't like completing the square if I don't have to

Answer 39

why do you think PRIME is not a correct response ? @v_nessa_87

Answer 40

because I submitted that as the answer and it returned incorrect

Answer 41

try submitting this
\[-t^2-9t+1 = -1(t^2+9t-1)\]

there is nothing much you can do here really..

Answer 42

(ofcourse in integers)

Answer 43

Oh man, I can't believe all it was was factoring out -1. If it had been t^2+9t-1, would it have been prime? That drives me nuts.

Answer 44

lol does that work ?

Answer 45

Yeah, it worked... *sigh*

Answer 46

yaaay! to me, t^2+9t-1 is as much prime as -t^2-9t+1 is prime
factoring out -1 is silly but yeah just give the grader whatever it wants haha!

Answer 47

oh jeeze, these online assignments are so dumb

Answer 48

^agreed

Answer 49

thanks all for your help!