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i would factor it, then make a table hen graph iy
almost.... you need to factor more
ok... you have a perfect square in the form of \[(a^2-b^2) = (a+b)(a-b) \]
\[(x^2-4) \] is a perfect square.. what is the square root of x^2 and what is the square root of 4
a perfect square is the product of a rational number multiplied by itself. the product of a polynomial multiplied by itself.
A perfect square is a number that can be expressed as the product of two equal integers.
what are you doing?
a is squared and so does x b is squared but 4 is the square of what number?
\[f(x) = x^3 + 4x^2 - x - 4\] factor by grouping which you did and that's fine you got to \[(x^2-4)(x+1) \] but you need to factor more..
\(\Large (a^2-b^2)=(a+b)(a-b)\) \(\Large (x^2-4)=(x^2-2^2)\) clear?
something's wrong here.
factor my grouping (at least when I did it) had something else as a perfect square
\[f(x) = x^3 + 4x^2 - x - 4 \rightarrow x^2(x+4)-1(x+4)\]
you shouldn't rearrange the terms.. -_- that's why you had a different answer.
second line is where everything got messed up x^3+x-4x^2-4 x(x^2+1)-4(x^2+1) you have an imaginary number case
Leave the function alone! *FIRST STEP* Second step use factor by grouping (two pairs!) first and second term has a common factor.. third and fourth term has a common factor
\[f(x) = x^3 + 4x^2 - x - 4 \] look at the first two terms.. what do they have in common?
ok .. we have four terms x^3 <- 1st term 4x^2 - 2nd term -x <- 3rd term -4 < - 4th term focus on the first and second only what do they have in common?
yes... how many x's though?
\[x^3-4x^2 => (x)(x)(x)-4(x)(x) \]
how many x's can I take out?
WHAT! I have three x's and I have 2 x's
-x-4 now what can I take out? =_=
what can I yank out to make sure I have (x+4) ?
I have -x-4 I want x+4 what do I need to yank out?