Factorise: $a^{2} + 8 a + 16$

(a - 4)^{2}

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

(a + 4)^{2}

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

(a + 4)

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

(a - 4)

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

Open in App

Solution

The correct option is B $(a + 4)^{2}$
$a^{2} + 8 a + 16$
$= a^{2} + 4 a + 4 a + 16$
$= a (a + 4) + 4 (a + 4)$
$= (a + 4) (a + 4)$
$= (a + 4)^{2}$

Suggest Corrections

Similar questions

Q. Factorise using identities:

(a^{2})^{2} + 2 \times a^{2} \times 4 + 4^{2}

Q. Factorise

a^{2} + 8 a + 16

Factorise:

$9 a^{2} - (a^{2} - 4)^{2}$

Factorise :

$(i) a^{2} - (b - c)^{2} (i i) 25 (2 x - y)^{2} - 16 (x - 2 y)^{2} (i i i) 16 (5 x + 4)^{2} - 9 (3 x - 2)^{2} (i v) 9 x^{2} - \frac{1}{16} (v) 25 (x - 2 y)^{2} - 4$

Factorise the following expressions :
$a^{2} + 8 a + 16$

Join BYJU'S Learning Program

Factorise: a2+8a+16

The correct option is B (a+4)2 a2+8a+16=a2+4a+4a+16=a(a+4)+4(a+4)=(a+4)(a+4)=(a+4)2

Factorise: $a^{2} + 8 a + 16$

The correct option is B $(a + 4)^{2}$
$a^{2} + 8 a + 16$
$= a^{2} + 4 a + 4 a + 16$
$= a (a + 4) + 4 (a + 4)$
$= (a + 4) (a + 4)$
$= (a + 4)^{2}$