Precalc factoring $x^2+10x+25-9y^2$

Question

Factor $$x^2+10x+25-9y^2$$

The solution is

$$(x+5-3y)(x+5+3y)$$

I understand how to factor when there is only one variable $x$ but I am not sure how to complete this problem with the additional variable $y$.

score 5 · Accepted Answer · answered Dec 02 '19 at 15:28

5

Consider $x$ only: $x^2+10x+25=(x+5)^2$, then you can use it with the whole expression:

$$x^2+10x+25 - 9y^2=(x+5)^2-(3y)^2=(x+5-3y)(x+5+3y)$$

Because $a^2-b^2=(a-b)(a+b)$, you only need to substitute $a=x+5$ and $b=3y$.

answered Dec 02 '19 at 15:28

Andronicus

I really appreciate your answer it makes it very clear, ill accept it when the timer allows. Thanks! – user123 Dec 02 '19 at 15:30
1

@user123 I'm very glad I could help:) – Andronicus Dec 02 '19 at 15:31
You can do factorization of general 2nd degree equation $ ax^2+2hxy+by^2+2gx+2fy+c=0$ if they represent pair of lines. – mathsdiscussion.com Dec 02 '19 at 15:58

score 2 · Answer 2 · answered Dec 02 '19 at 15:29

2

Note that $$a^2-b^2=(a-b)(a+b)$$ Now put that $$a=x+5$$ $$b=3y$$

answered Dec 02 '19 at 15:29

ajotatxe

score 0 · Answer 3 · answered Dec 02 '19 at 16:02

0

The given equation can be treated like quadratic in y assuming x to be constant.

Then y=$\frac{\pm \sqrt{36(x^2+10x+25)}}{18}$ $$3y=(x+5) \,\,or\,\,3y=-(x+5)$$

answered Dec 02 '19 at 16:02

This is missing a final step. You have found the roots of the expression, which says the relationship between $x$ and $y$ when it equals $0$. But this does not mean that the original equation is equal to $(3y-(x+5))(3y+(x+5))$ or even $(y-r_1)(y-r_2)$ for roots $r_i$. This is because a quadratic is only determined by its roots up to the coefficient of $x^2$. In our case, the coefficient is $-9$ so we'd need to flip the sign: $-(3y-(x+5))(3y+(x+5))$. – Jam Dec 02 '19 at 17:16
I am talking in general way to solve such types of 2nd degree equations of the form $$ax^2+2hxy+by^2+2gx+2fy+c=0 $$ if expression is not equal to zero then always we can write poly. As $$p(x)=a(x-x_1)(x-x_2)...(x-x_n)$$ where $$x_1,x_2,...,x_n$$ are roots of equation p(x)=0. If you wanted only particular solution to your question not the general approach for such question then really I am very sorry. – mathsdiscussion.com Dec 02 '19 at 17:21
I understand that, I'm saying you need to specify the coefficient of the highest degree variable to determine the polynomial. See Question 3253240 - polynomials are not uniquely determined by their roots. – Jam Dec 02 '19 at 17:23

3 Answers3