Factorise $(2 x^{2} + 13 x + 15)$

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

$(2 x - 3) (x - 5)$

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

$(2 x + 3) (x + 10)$

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

$(2 x + 3) (2 x + 5)$

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

Open in App

Solution

The correct option is A
5

$2 x^{2} + 13 x + 15$

The product of the first and last term = $15 \times 2 x^{2} = 30 x^{2}$
and the sum or difference = 13 x ;

Hence $30 x^{2} = 10 x \times 3 x$

= $2 x^{2} + 10 x + 3 x + 15$
= $2 x (x + 5) + 3 (x + 5)$
= $(2 x + 3) (x + 5)$

Suggest Corrections

Similar questions

Factorise $2 x^{2} + 13 x + 15$

Q. Factorise:

2 x^{2} - 7 x - 15

Q. Solve the following equation:

2 x^{2} + 13 x + 15 = 0

Factorise the following:

2 x^{2} - 7 x - 15

Using the remainder Theorem, factorise each of the following completely:
(i) $3 x^{3} + 2 x^{2} - 19 x + 6$
(ii) $2 x^{3} + x^{2} - 13 x + 6$
(iii) $3 x^{3} + 2 x^{2} - 23 x - 30$
(iv) $4 x^{3} + 7 x^{2} - 36 x - 63$
(v) $x^{3} + x^{2} - 4 x - 4.$

Join BYJU'S Learning Program

Factorise (2x2+13x+15)

The correct option is A 5 2x2+13x+15 The product of the first and last term = 15×2x2=30x2 and the sum or difference = 13 x ; Hence 30x2=10x×3x = 2x2+10x+3x+15 = 2x(x+5)+3(x+5) = (2x+3)(x+5)

Factorise $(2 x^{2} + 13 x + 15)$

The correct option is A
5

$2 x^{2} + 13 x + 15$

The product of the first and last term = $15 \times 2 x^{2} = 30 x^{2}$
and the sum or difference = 13 x ;

Hence $30 x^{2} = 10 x \times 3 x$

= $2 x^{2} + 10 x + 3 x + 15$
= $2 x (x + 5) + 3 (x + 5)$
= $(2 x + 3) (x + 5)$