Factorise:
$p^{2} - 3 p q + 2 q^{2}$

(3 p - 2 q) (p + q)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(3 p - 2 q) (p - q)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(p - 2 q) (p - q)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(p - q) (p - q)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $(p - 2 q) (p - q)$
$p^{2} - 3 p q + 2 q^{2}$
$= p^{2} - 2 p q - p q + 2 q^{2}$
$= p (p - 2 q) - q (p - 2 q)$
$= (p - 2 q) (p - q)$

Suggest Corrections

Similar questions

$(p + 2 q) (p - q)$ = $p^{2} + p q$ - _____

(p + 2 q) (p + 3 q)

p = 1

q = 0

2 q (3 p^{2} - 3 p q + 8) - 3 p (p - q)

Factorise: p^{2} + q^{2} + 2 (p q + q r + r p) .

(3 p + q, p - q) = (p - q, 3 p + q)

Join BYJU'S Learning Program