Factorize $5 - 20 x^{2}$ is

5 (1 + 2 x) (1 + 2 x)

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

5 (1 + 2 x) (1 - 2 x)

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

(1 + 3 x) (1 - 3 x)

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

None

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

Open in App

Solution

The correct option is C $5 (1 + 2 x) (1 - 2 x)$
Factorization of $5 - 20 x^{2} .$
$\Rightarrow$ Take $5$ common from the expression, $5 (1 - {4 x}^{2}) = 5 (1^{2} - {(2 x)}^{2})$
$\Rightarrow$ Using $a^{2} - b^{2} = (a + b) (a - b)$
$= 5 (1 - 2 x) (1 + 2 x) .$
Hence, the answer is $5 (1 - 2 x) (1 + 2 x) .$

Suggest Corrections

Similar questions

n

\frac{1}{(1 + x) (1 + 2 x)} + \frac{1}{(1 + 2 x) (1 + 3 x)} + \frac{1}{(1 + 3 x) (1 + 4 x)} + . . . . .

n

\frac{1}{(1 + x) (1 + 2 x)} + \frac{1}{(1 + 2 x) (1 + 3 x)} + \frac{1}{(1 + 3 x) (1 + 4 x)} + . . .

Sum to n terms of the series $\frac{1}{(1 + x) (1 + 2 x)}$ + $\frac{1}{(1 + 2 x) (1 + 3 x)}$ + $\frac{1}{(1 + 3 x) (1 + 4 x)}$ + ..............is

$(2 x - 32 x^{3}) = ? (a) 2 (x - 4) (x + 4) (b) 2 x (1 - 2 x)^{2} (c) 2 x (1 + 2 x)^{2} (d) 2 x (1 - 4 x) (1 + 4 x)$

\frac{(1 + x) (1 + 2 x) (1 + 3 x)}{(1 - x) (1 - 2 x) (1 - 3 x)} = K +

\frac{A}{1 - x} + \frac{B}{1 - 2 x} + \frac{C}{1 - 3 x}

Join BYJU'S Learning Program