Factorise n2-22 n-96A- n-16n-6B- n-16n-6C- n-16n-6D- n-16n-6

The correct option is A (n + 16)(n + 6)(n^2 + 22n + 96) = n^2 + 16n + 6n + 96 = n(n + 16) + 6n(n + 16) = (n + 6)(n + 16) ...

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

Division of a Polynomial by a Monomial

Factorise n2+...

Question

Factorise $(n^{2} + 22 n + 96)$

(n – 16)(n + 6)

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

(n + 16)(n – 6)

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

(n + 16)(n + 6)

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

(n – 16)(n – 6)

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

Open in App

Solution

Suggest Corrections

Similar questions

n^{2} - 16 n - 59 = 0

(n - 6)^{2} + (\frac{1}{(n - 6)^{2}}) = ?

49^{n} + 16 n - 1

n \in N; 49^{n} + 16 n - 1

16^{n + 1} = 64 \times 4^{- n}

n

A = {\begin{matrix} 5^{n} - 4 n - 1 : n \in N \end{matrix}}

B = {\begin{matrix} 16 (n - 1) : n \in N \end{matrix}}

Join BYJU'S Learning Program