Question 12 Marks
Factorise $15 x^2 y+6 x^4 y^2-9 x y$ and find its irreducible form.
Answer
View full question & answer→We can write
$
\begin{aligned}
15 x^2 y & =3 \times 5 \times x \times x \times y \\
6 x^4 y^2 & =2 \times 3 \times x \times x \times x \times x \times y \times y \\
9 x y & =3 \times 3 \times x \times y
\end{aligned}
$
Here, $3 \times x \times y$ is common in these terms.
$
\begin{aligned}
\therefore 15 x^2 y+ & 6 x^4 y^2-9 x y \\
& =3 \times 5 \times x \times x \times y+2 \times 3 \times x \times x \times x \times x \times y \times y \\
& \quad-3 \times 3 \times x \times y \\
& =3 \times x \times y[5 \times x+2 \times x \times x \times x \times y-3 \times 1] \\
& =3 x y\left(5 x+2 x^3 y-3\right)
\end{aligned}
$
$
\begin{aligned}
15 x^2 y & =3 \times 5 \times x \times x \times y \\
6 x^4 y^2 & =2 \times 3 \times x \times x \times x \times x \times y \times y \\
9 x y & =3 \times 3 \times x \times y
\end{aligned}
$
Here, $3 \times x \times y$ is common in these terms.
$
\begin{aligned}
\therefore 15 x^2 y+ & 6 x^4 y^2-9 x y \\
& =3 \times 5 \times x \times x \times y+2 \times 3 \times x \times x \times x \times x \times y \times y \\
& \quad-3 \times 3 \times x \times y \\
& =3 \times x \times y[5 \times x+2 \times x \times x \times x \times y-3 \times 1] \\
& =3 x y\left(5 x+2 x^3 y-3\right)
\end{aligned}
$

