Question

Factor the trinomial: x2 - 14x + 24
A. (x - 3)(x - 8)
B. (x - 2)(x - 12)
C. (x - 4)(x + 12)
D. prime

Answer 1

I get B im not sure tho

Answer 2

@zepdrix

Answer 3

So we want `factors` of positive 24 that will also `add` to -14.

Let's check B a sec.
The given factors are -2 and -12.
\[\Large -2\times-12=24\]\[\Large -2+-12=-14\]

Those seem to be the correct values that we want! :)

Answer 4

its -12 x -2

Answer 5

\[x^{2}-12x-2x+24 \]

Answer 6

\[(x ^{2}-12)(-2x+24)\]

Answer 7

@zepdrix \[x(x-12)2x(-1+24)?\]

Answer 8

Oh you're trying to see the factors by grouping?
Start by factoring an x out of the first two terms.
\[\Large x^{2}-12x-2x+24 \quad=\quad x(x-12)-2x+24\]
Then factor a -2 out of each of the second terms.\[\Large x\color{royalblue}{(x-12)}-2\color{royalblue}{(x-12)}\]

Answer 9

Soo im corecct
B

Answer 10

\[(x-12)(x-2)\]

Answer 11

1 Question what would it have to be in order for it to be prime ?i don't get prime.

Answer 12

@zepdrix

Answer 13

oh prime? :o ummmm

Answer 14

Yeah idk that

Answer 15

Think about what it means for a `number` to be prime.
A number is prime if it's only factors are 1 and itself, right?

Example: 7 has factors 1x7, but no other factors.
So it is a prime number.

Answer 16

For a `polynomial` to be prime, it can only be written as the product of 1 and itself.\[\Large x^2-2x+7 \quad=\quad (1)(x^2-2x+7)\]

Answer 17

In the example above, there is no way for us to write x^2-2x+7 as any other factors. (at least not with real integer values).\[\Large x^2-2x+7 \quad \ne \quad(\qquad)(\qquad)\]

Answer 18

Maybe I'm making this more complicated than it needs to be :) lol
It simply means there is no way for us to factor the polynomial.

Answer 19

NO no i get it thanks