1. Originally Posted by jay123
lol...im still lost.

do i
X^2-5(2y+3)+6
X^2-10y-15+6
X^2-10y+1
you already had (x - 2)(x - 3), why would you go back to the expanded form? just replace ALL x's in the foiled form with 2y + 3. that's it

2. sorry. but do replace all (2y+3) with x.?

3. Originally Posted by jay123
sorry. but do replace all (2y+3) with x.?
?? i'm sorry, but i don't see your hang up here

$({\color {red} x} - 2)({\color {red}x} - 3) = ({\color {red} 2y + 3} - 2)( {\color {red} 2y + 3} - 3)$

now simplify

This is my 37th post!!!

4. (x-3)(x-2) (2y+3-2)(2y-3-3) = 2y+1 2y = 2y(2y+1). yessss, thank you..but how do i recognize this as a quad. equation

5. Originally Posted by jay123
(x-3)(x-2) (2y+3-2)(2y-3-3) = 2y+1 2y = 2y(2y+1). yessss, thank you..but how do i recognize this as a quad. equation
quadratics are of the form: $ax^2 + bx + c$

you had something squared plus a constant times something plus a constant, that is in the form of the quadratic above. in this case you had $a = 1$, $x = 2y + 3$, $b = -5$ and $c = 6$. so they were in that form

and it was not an equation, you need an equal sign to be an equation

-3x^3-15x^2+27x+135
i first grouped: (-3x^3-15x^2+27x)+135 = -3x(x^2+5x+9)+135

so i decided to regroup the 5x+135 = 5(x+27)

-3x(x^29) = -3x(x+3)(x-3)5(x+27)

now do i -3x+5 (x+3)(x+27)?

7. Originally Posted by jay123

-3x^3-15x^2+27x+135
i first grouped: (-3x^3-15x^2+27x)+135 = -3x(x^2+5x+9)+135

so i decided to regroup the 5x+135 = 5(x+27)

-3x(x^29) = -3x(x+3)(x-3)5(x+27)

now do i -3x+5 (x+3)(x+27)?
this is the same type of question as the first one you asked. there is a common $-3x^2$ among the first two terms and a common 27 among the last two. factor by grouping

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