17 Factorise completely $$ x^{2}+2 x y+y^{2}+3 x+3 y-18 $$

Question

UpStudy AI · Accepted Answer

We start with the expression

$$
x^{2}+2xy+y^{2}+3x+3y-18.
$$

Notice that the first three terms form a perfect square:

$$
x^{2}+2xy+y^{2} = (x+y)^2.
$$

Thus, we can rewrite the expression as:

$$
(x+y)^2 + 3x + 3y - 18.
$$

Factor $$3x+3y$$ as $$3(x+y)$$:

$$
(x+y)^2 + 3(x+y) - 18.
$$

Let $$ t = x+y $$. The expression becomes:

$$
t^2 + 3t - 18.
$$

We factor the quadratic in $$ t $$ by finding two numbers that multiply to $$-18$$ and add to $$3$$. The numbers $$6$$ and $$-3$$ satisfy this, since

$$
6 \times (-3) = -18 \quad \text{and} \quad 6 + (-3) = 3.
$$

Thus, we factor the quadratic:

$$
t^2+3t-18 = (t+6)(t-3).
$$

Substitute back $$ t = x+y $$:

$$
(x+y+6)(x+y-3).
$$

Therefore, the completely factorised form is:

$$
(x+y+6)(x+y-3).
$$
The expression $$ x^{2}+2xy+y^{2}+3x+3y-18 $$ can be factored completely as $$ (x+y+6)(x+y-3) $$.

17 Factorise completely