If $p (x) = x^{2} + 6 x + 8, q (x) = x^{2} + 5 x + 6, r (x) = x^{2} + 4 x + 3$ and $s (x) = x^{2} + 5 x + 4$ then choose the correct options -

p (x) . s (x) = q (x) . r (x)

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

p (x) . s (x) \neq q (x) . r (x)

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

p (x) . q (x) = s (x) . r (x)

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

N o n e o f t h e s e

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

Open in App

Solution

The correct option is B $p (x) . s (x) \neq q (x) . r (x)$
If $p (x) = x^{2} + 6 x + 8 = (x + 4) (x + 2)$
$q (x) = x^{2} + 5 x + 6 = (x + 3) (x + 2)$
$r (x) = x^{2} + 4 x + 3 = (x + 3) (x + 1)$
$s (x) = x^{2} + 5 x + 4 = (x + 4) (x + 1)$
Now, $p (x) . s (x) = (x + 4) (x + 2) \times (x + 4) (x + 1)$ .......... (1)
$q (x) . r (x) = (x + 3) (x + 2) \times (x + 3) (x + 1)$ .............. (2)
From eq. (1) and (2), $p (x) . s (x) \neq q (x) . r (x)$

Suggest Corrections

Similar questions

p (x) = x^{2} - 4, q (x) = x^{3} - 8, r (x) = (x + 2)

s (x) = (x^{2} + 2 x + 4)

P (x) + q (x) = x^{2} - 4 x + 1 a n d P (x) - q (x) = x^{2} + 5 x - 2

p (x) = x^{2} - 5 x + 6

q (x) = x^{2} + 5 x + 4

p (x) = x^{2} - 2 x + 1

q (x) = x^{4} - 1

r (x) = x^{3} - 2 x^{2} - 5 x + 6

Now for each pair of polynomials p(x) and q(x) given below, find p(x) + q(x) and p(x) − q(x). Also compute p(1) + q(1) and p(1) − q(1) for each.

(i)

(ii)

(iii)

(iv)

(v)

Join BYJU'S Learning Program