Determine the HCF of $a^{2} - 25, a^{2} - 2 a - 35$ and $a^{2} + 12 a + 35$

(a-5)(a+7)

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

(a+5)(a-7)

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

(a-7)

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

(a+5)

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

Open in App

Solution

The correct option is D (a+5)
Since, $a^{2} - 25 = (a - 5) (a + 5)$
$a^{2} - 2 a - 35 = a^{2} - 7 a + 5 a - 35$
$= a (a - 7) + 5 (a - 7)$
$= (a + 5) (a - 7)$
and
$a^{2} + 12 a + 35 = a^{2} + 7 a + 5 a + 35$
$= (a + 7) (a + 5)$
Clearly HCF of $a^{2} - 25, a^{2} - 2 a - 35$ and $a^{2} + 12 a + 35$ i.e $(a - 5) (a + 5), (a + 5) (a - 7)$ and $(a + 7) (a + 5)$ is $a + 5$
Option D is correct.

Suggest Corrections

Similar questions

5 a^{2} - 3 a + 2

- 8 a^{2} + 5 a + 6

a^{2} + 7 a - 4

Divide (a^{2} + 7 a + 10) by (a + 5) .

a^{2} + b^{2} = 34

a b = 12

7 (a - b)^{2} - 2 (a + b)^{2}

\frac{cos 2 A cos 3 A - cos 2 A cos 7 A + cos A cos 10 A}{sin 4 A sin 3 A - sin 2 A sin 5 A + sin 4 A sin 7 A} = cot 6 A cot 5 A

p (a) = 2 a^{2} - 7 a + 3

q (a) = 2 a^{2} + 5 a - 3

Join BYJU'S Learning Program