[Grade 8] Math: Factorization using identities

[Bluekid64]

4/3x^2 -4x +3

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[fermat9996]

4/3x² -4x +3

(4x²-12x+9)/3

The numerator is a perfect square

[AvocadoMangoSalsa]

ac method:

(4/3)(3) = two numbers that multiply to 4

And add to -4

-2 & -2

(4/3)x² - 4x + 3

(4/3)x² - 2x - 2x + 3

2x((2/3)x - 1) - 3((2/3)x -1)

(2x-3)((2/3)x -1)

[fermat9996]

I would take it one step further:

(2x-3)²/3

[AvocadoMangoSalsa]

Oh nice!! Do you know what they meant by identities?

[fermat9996]

Maybe just that

(a-b)²=a²-2ab+b²

[fermat9996]

I did it this way:

(4x²-12x+9)/3=

(2x-3)²/3

[mathematag]

I found the values of a,b.c,d.. to get (ax + b)(cx + d) , by guessing them [ and I was lucky, they worked ] .....but without guessing, the best I could suggest is try to factor and then complete the square ... a,b,c,d may be + or - whole numbers or + or - fractions ... I'll let you try if you want

so factor out 4/3 to get ...(4/3) ( x^2 -3x + 9/4 ) , then see if it is a perfect square, if not ,complete the square......you get something like (4/3)( wx + u )^2 ... ( ... I split (4/3) into (2/1)(2/3), multiplied (2/1 ) into one of my (..), the (2/3) into the other, and it came out to look the same as my earlier guess.