algebra-precalculusfactoringpolynomials
I've been working on this for hours and cannot figure it out.
When I search, I find factorization techniques that I already know but don't seem to be able to apply here, or that are for polynomials that don't have the same form.
I am beginning to wonder if this can even be factored.
$$4x^2 + 4x – 9y^2 -1$$
The most I can figure to factor is:
$$4x(x+1) – 9y^2 -1$$
Mahalo for the help, I am really trying to understand this.
With the polynomial $x^4 -7x^3 + 3x^2 + 31x + 20$ where $(x+1)$ is a factor for example.
Do I use the same procedure but start a degree higher, like $(x+1)(\alpha x^3 + \beta x^2 + \gamma x + \delta)$ and move on down to figure out the 2nd degree after that?
There are other techniques, namely the polynomial long division as mentioned in a comment by pedja, the Ruffini's rule and for polynomials with integer coefficients the rational root theorem, but you can apply the method of equating coefficients to
$$\begin{eqnarray*} P(x) &:&=x^{4}-7x^{3}+3x^{2}+31x+20 =\left( x+1\right) \left( \alpha x^{3}+\beta x^{2}+\gamma x+\delta \right). \end{eqnarray*}\;\; \tag{1}$$
Expanding the RHS and comparing with the LHS, where the coefficient of $x^{4}$ is $1$, you would conclude that $\alpha =1$ and are left with a simpler equality
$$\begin{eqnarray*} P(x) &:&=x^{4}-7x^{3}+3x^{2}+31x+20=\left( x+1\right) \left( x^{3}+\beta x^{2}+\gamma x+\delta \right) \\ &=&x^{4}+\left( \beta +1\right) x^{3}+\left( \beta +\gamma \right) x^{2}+\left( \gamma +\delta \right) x+\delta . \end{eqnarray*}$$ Equating coefficients you get the simple system of four linear equations $$ \begin{equation*} \left\{ \begin{array}{c} \delta =20 \\ \gamma +\delta =31 \\ \beta +\gamma =3 \\ \beta +1=-7, \end{array} \right. \end{equation*}$$ whose solution is $\delta =20,\beta =-8,\gamma =11$. So
$$\begin{equation*} x^{4}-7x^{3}+3x^{2}+31x+20=\left( x+1\right) \left( x^{3}-8x^{2}+11x+20\right). \end{equation*}\tag{2}$$ Now by direct inspection (or by the rational root theorem, since $1,2,4$ and $5$ are the positive divisors of $20$) you can find that $x=-1$ is a zero of the cubic polynomial $Q(x):=x^{3}-8x^{2}+11x+20$, ie $Q(-1)=0$. Applying a similar method to $Q(x)$ you would find that
$$\begin{equation*} Q(x)=x^{3}-8x^{2}+11x+20=\left( x+1\right) \left( x-4\right) \left( x-5\right) . \end{equation*}$$
Consequently, $$\begin{equation*} x^{4}-7x^{3}+3x^{2}+31x+20=\left( x+1\right) ^{2}\left( x-4\right) \left( x-5\right). \end{equation*}\tag{3}$$
A shorter method is the Ruffini's rule applied to the polynomial division of $P(x)$ by $(x-r)$. This case (with $P(x)=x^{4}-7x^{3}+3x^{2}+31x+20$ and $r=-1,P(r)=0$, ) is shown bellow, where the remainder is $P(r)=s$.
How wide is the shaded region? If the length of the larger rectangle is $a^3$, and the length of the right is $b^3$...
The width of the shaded region is $a^3-b^3$
How tall is the shaded region?
I assume the picture is indicating that the height is $a^2+ab+b^2$
What is the area of a rectangle described as in terms of width and height?
width times height
And what is that in this case?
$(a^3-b^3)(a^2+ab+b^2)$
What is the perimeter of a rectangle described as in terms of width and height?
twice the sum of width and height
And what is that in this case?
$2(a^3-b^3 + a^2+ab+b^2)$
Now... can we express any of these above in a more compact way via factoring? That is debatable, but I do recognize at least one simplification you can do for area:
Knowing that $a^3-b^3=(a-b)(a^2+ab+b^2)$ we have that the area can be described as $(a-b)(a^2+ab+b^2)^2$. Is that really more preferred than $(a^3-b^3)(a^2+ab+b^2)$? I don't see a big difference between the two, so personally I wouldn't care which. They are the same to me.
Using the same for perimeter, we can do the following:
$2(a^3-b^3 + a^2+ab+b^2) = 2((a-b)(a^2+ab+b^2)+1(a^2+ab+b^2)) = 2(a-b+1)(a^2+ab+b^2)$. Again, is this more useful than the first way of writing it? That is debatable...They are not fundamentally different.
Best Answer
While the expression cannot be factored (it cannot be represented strictly as the product of factors, we can proceed to manipulate it in a way that involves two factors, with an added (subtracted) constant: $$ \begin{align} 4x^2 + 4x - 9y^2 - 1 &= 4x^2 + 4x + \color{red}{1} - 9y^2 - 1 \color{red}{-1} \\ &= (2x+1)^2 - (9y^2 + 2) \\ &= (2x+1)^2 - (3y)^2 -2 \\ &= (2x + 1 +3y)(2x + 1 - 3y) -2 \\ &= (2x + 3y +1)(2x - 3y + 1) -2 \;? \end{align} $$
I'm "simply" simplifying the expression you gave (which is not an equation). If you meant $$4x^2 + 4x - 9y^2 - 1 = 0$$ you could write:
$$4x^2 + 4x - 9y^2 - 1=0 $$ $$ \iff (2x+1)^2 - (3y)^2 = 2 $$ $$\iff (2x + 1 +3y)(2x + 1 - 3y) = 2 $$ $$\iff (2x + 3y +1)(2x - 3y + 1) = 2$$