Question

factorise the following expression if possible by completing the square:
3x^2 -6x-5

Answer 1

First, you need to take 3 as common, as the "a" value needs to be 1
Rewrite Equation:
\[3(x^2-2x+\frac{-5}{-4})\]
Simplify:
\[3(x^2-2x+\frac{5}{4})\]
To compelete the square, you need to first do this:
\[(\frac{b}{2})^2\]
Sub in the b value
\[(\frac{1}{2})^2\]
This gives:
\[0.25\]
Then put this value between the "b" and "c" value, making sure to take away
\[3(x^2-2x+0.25-0.25+\frac{5}{4})\]
You can group the first 3 values
\[3((x-0.25)^2-0.25+\frac{5}{4})\]
Simplify
\[3((x-0.25)^2+1)\]
Expand:
\[3(x-0.25)^2+3\]
Square root, then square the 3
\[3(x-0.25)^2+(\sqrt{3})^2\]
Then use the Difference of Square Law
\[3(x-0.25-\sqrt{3})(x-0.25+\sqrt{3})\]
Done, phew ;)

Answer 2

@TL9800 Hope this helped. Since this is such a winded question, need someone like @iambatman to clarrify

Answer 3

mhm

Answer 4

Whoops, replace the 4 by 3. Not sure where that came from ;)

Answer 5

\[3x^2-6x-5\]

you can group the terms containing x, leaving a blank spot to add constant in later

\[(3x^2-6x+~~~)-5 \]

I think you forgot to do this part @Ahsome

factor of a 3 now

\[3(x^2-2x+~~~~)-5\]

\[3(\frac{ -2 }{ 2 })^2 =3\]

\[3(x^2-2x+1)-5-3 \implies 3(x^2-2x+1)-8 \implies 3(x-1)^2-8\]

Answer 6

That's another way. This was the way I learnt in school. @TL9800 Remember to change 4 to 3, error there ;)

Answer 7

thank uoi

Answer 8

No Problem mate ;)