Question 89 (xi)

Factorise the following, using the identity, $a^{2} + 2 a b + b^{2} = (a + b)^{2}$

$9 x^{2} + 24 x + 16$

Open in App

Solution

We have, $9 x^{2} + 24 x + 16$
$= (3 x)^{2} + 2.3 x .4 + 4^{2} = (3 x + 4)^{2} = (3 x + 4) (3 x + 4)$

Suggest Corrections

Similar questions

a^{2} + 2 a b + b^{2} = (a + b)^{2}

9 x^{2} + 30 x + 25

Factorise the following, using the identity, $a^{2} + 2 a b + b^{2} = (a + b)^{2}$ :

$9 x^{2} + 2 x y + \frac{y^{2}}{9}$

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

a^{2} y^{3} - 2 a b y^{2} + b^{2} y

a^{2} + 2 a b + b^{2} = (a + b)^{2}

2 x^{3} + 24 x^{2} + 72 x

a^{2} + 2 a b + b^{2} = (a + b)^{2}

a^{2} x^{3} + 2 a b x^{2} + b^{2} x

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program