Factorize : $p^{2} + 6 p - 16$

(p - 8) (p - 2)

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

(p + 8) (p - 2)

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

(p - 6) (p - 2)

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

(p + 6) (p + 2)

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

Open in App

Solution

The correct option is D $(p + 8) (p - 2)$
$p^{2} + 6 p - 16$
$= p^{2} + 8 p - 2 p - 16$
$= p (p + 8) - 2 (p + 8)$
$= (p + 8) (p - 2)$

Suggest Corrections

Similar questions

2 p^{3} - 3 p^{2} + 4 p - 5 - 6 p^{3} + 2 p^{2} - 8 p - 2 + 6 p + 8

p^{2} + 6 p + 8

q^{2} - 10 q + 21

p^{2} + 6 p - 16

p^{2} + 6 p - 16

p^{2} + 6 p - 16

(i) p^{2} + 6 p + 8 (i i) p^{2} - 10 q + 21 (i i i) p^{2} + 6 p - 16

Join BYJU'S Learning Program