MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Factorising a Binomial

Factorise:2 x...

Question

Factorise:
$2 x p + 4 x q - 3 y p - 6 y q$

A

$(p + 2 q) (2 x - 3 y)$

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

B

$(p - 2 q) (2 x - 3 y)$

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

C

$(p + 2 q) (2 x + 3 y)$

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

D

$(p - 2 q) (2 x + 3 y)$

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

Open in App

Solution

The correct option is A
$(p + 2 q) (2 x - 3 y)$

The expression $2 x p + 4 x q - 3 y p - 6 y q$ can be regrouped as,

$(2 x p + 4 x q) + (- 3 y p - 6 y q)$

Taking $2 x$ and $- 3 y$ common, we get

$= 2 x (p + 2 q) - 3 y (p + 2 q)$

Taking $p + 2 q$ cmmon, we get
$= (p + 2 q) (2 x - 3 y)$

$\Rightarrow$ $2 x p + 4 x q - 3 y p - 6 y q$ $= (p + 2 q) (2 x - 3 y)$

Therefore, the factors of $2 x p + 4 x q - 3 y p - 6 y q$ are $(p + 2 q)$ and $(2 x - 3 y)$ .

Suggest Corrections

thumbs-up

0

Similar questions

2 x p + 4 x q - 3 y p - 6 y q = (p + 2 q) \times

Factorise $1 - (2 x - 3 y)^{2}$

Q. Factorise the expression 4x + 12xy.

Q. Factorise the expression:

50 x^{3} - 18 x y^{2}

2 x^{2} - 18 y^{2}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Factorising a Binomial

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Factorising a Binomial

Standard VII Mathematics

Join BYJU'S Learning Program

CrossIcon