factor each completely
a^2+6a-16

Question

MAGABACK · Answer

is that a to the power of 2?

HappyXTheXBunny · Answer

yessir

MAGABACK · Answer

so is the equation$$y=a^2+6a-16$$

MAGABACK · Answer

cause I think it would need a y= in it

HappyXTheXBunny · Answer

idk thats why im asking💀

HappyXTheXBunny · Answer

like bfr gang💀

MAGABACK · Answer

It might be like this to set the equation up. $$a^2+6a-16=0$$

MAGABACK · Answer

now u need to figure out what =-16 but equals 6

MAGABACK · Answer

@bot1 wrote:

dude i gave the answer already

ur not supposed to give a direct answer

0mega · Answer

@bot1 wrote:

a^2 +6a−16=(a+8)(a−2) is the final factorization

you can 't give a direct answer

MAGABACK · Answer

and I think the answer is wrong anyway

MAGABACK · Answer

8x-2=-16 but nvm. -2+8 would be 6, so ig its right

MAGABACK · Answer

but next time plz don't tell users direct answers

HappyXTheXBunny · Answer

tx bot

toga · Answer

@magaback wrote:

but next time plz don't tell users direct answers

and if you gonna at least explain how you got it

Extrinix · Answer

You don't need a "$y=$" nor a "$=0$" to factor something. $a^2+6a-16$ Since we can't take out a portion entirely, what factors can get you to 6a that also work with 16? (6 and 1? no) (8 and 2? yes) So we have $8$ and $2$, if we factor those out within the equation, we understand that the $a^2 \pm ba \pm c$ equation almost always ends up factoring out to $(a\pm m)(a \pm n)$ There fore, we can use the values we got above ($8$ and $2$) to create the factored-out form of our original equation, and plug them into our $(a\pm m)(a \pm n)$ equation. Be wary, you need to find whether the values are plus or minus.