$16 x + 4 z$ times $(- 2 y - 8 x)$ is equal to

- 128 x^{2} - 32 x y - 32 x z - 8 y z

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

- 128 y^{2} - 32 x y - 32 x z - 8 y z

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

Open in App

Solution

The correct option is A $- 128 x^{2} - 32 x y - 32 x z - 8 y z$
Applying First Out Inside Last (FOIL) technique we get,
$(16 x + 4 z) \times (- 2 y - 8 x)$
$= (16 x + 4 z) \times - (2 y + 8 x)$
$= - (16 x + 4 z) \times (2 y + 8 x)$
$= - (16 x \times 2 y + 16 x \times 8 x + 4 z \times 2 y + 4 z \times 8 x)$
$= - 32 x y - 128 x^{2} - 8 y z - 32 z x$
$= - 128 x^{2} - 32 x y - 32 x z - 8 y z$

Suggest Corrections

Similar questions

4 x^{3} y^{4} z^{5}

16 x^{2} y^{2} z^{2}

\int sin x . cos x . cos 2 x . cos 4 x . cos 8 x . cos 16 x d x

∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \begin{matrix} 4 x & 6 X + 2 & 8 X + 1 6 X & 9 X + 3 & 12 X 8 X & 12 X & 16 X + 2 \end{matrix} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣

\int (sin x cos x cos 2 x cos 4 x cos 8 x cos 16 x) d x

\int sin x . cos x . cos 2 x . cos 4 x . cos 8 x . cos 16 x d x

Join BYJU'S Learning Program