Factor completely: 32x^2 − 80x + 50
A. 2(4x − 5)(4x − 5) B. 2(4x + 5)(4x + 5) C. 2(8x − 5)(8x − 5) D. 2(8x + 5)(8x + 5)
Can you divide the whole equation by 2 first?
\(\color{#0cbb34}{\text{Originally Posted by}}\) @snowflake0531 Can you divide the whole equation by 2 first? \(\color{#0cbb34}{\text{End of Quote}}\) yes?
So what would you get
\[\frac{ 32x^2-80x+50 }{ 2 }\]
im confused :(
We can expand on ^^ equation to \[\frac{ 32x^2 }{ 2 } - \frac{ 80x }{ 2 } + \frac{ 50 }{ 2 }\]
ok, 16x, 40x and 25?
Yes so we have \[16x^2 -40x + 25\]
Now, which ones can you eliminate looking at 16x^2
hint, look at the choices where the differences are 4 and 8
common factor 32x-80+50
we can eliminate a and b?
No, if we eliminated a and b, we would be left with c and d, which would maek 64x^2
2(16X-40+25
ohhhh
Use the sum-product pattern Common factor from the two pairs
So you eliminate the third and fourth
\(\color{#0cbb34}{\text{Originally Posted by}}\) @snowflake0531 So you eliminate the third and fourth \(\color{#0cbb34}{\text{End of Quote}}\) ok
And so now, we just look at the middle one In this case, it is a negative number So, which one would we choose
2(4𝑥−5)^2
It's A
i think it's A
\(\color{#0cbb34}{\text{Originally Posted by}}\) @carmelle It's A \(\color{#0cbb34}{\text{End of Quote}}\) yes
tysm, im so bad at factoring o-o
tbh, I usually just like to guess around xd, and not actually think thru it
\(\color{#0cbb34}{\text{Originally Posted by}}\) @snowflake0531 tbh, I usually just like to guess around xd, and not actually think thru it \(\color{#0cbb34}{\text{End of Quote}}\) lol, I'll try that
So, especially with these, you have choices, you could just factor it out using FOIL and see which one is right
ok :D
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