Question

Carry out the multiple of the expreession ofbthe following pair pq+qr+rp +1,0

Answer 1

Answer:

v)  pq + qr + rp, 0

(pq + qr + rp) x 0 = (pq x 0) + (qr x 0) + (rp x 0)

= 0

Step-by-step explanation:

I think it is helpfull

Answer 2

$\huge\star\underline\mathfrak\pink{Solution}\star$

➡Algebraic expressions:

A combination of constants and variables connected by any or all of the four fundamental operations +, -,×,÷ is called an algebraic expression.

➡Terms:

The different parts of the expression separated by the sign+ or – are called the terms of the expression.

➡Monomials:

A monomial is a one term algebraic expression .

Multiplication of algebraic expression:

The product of two factors with like signs is positive and the product of two factors with unlike signs is negative.

Multiplication of a monomial and the binomial:

If P, q & r are the three monomials we use the distributive law.

p×(q+r)= p×q+ p×r

▅▅▅▅▅▅▅▅▅▅▅▅▅▅▅▅▅▅▅▅⠀⠀⠀⠀⠀⠀Solution:

❥..i)4p, q + r

4p(q + r) =  (4p x q) + (4p x r )

=4pq + 4pr

❥..ii)ab, a − b 

ab(a - b) =  = (ab x a) +  (ab x -a)

❥...=a²b - ab²

❥..iii) a + b, 7a²b²

(a + b) (7a²b²) = ( a x 7a²b²) + (b x 7a²b²)

❥...=7a³b²+ 7a²b³

❥..iv) a²− 9, 4a

(a² - 9)(4a) = (a² x 4a) + ( -9 x 4a)

= 4a³ - 36a

❥..v)  pq + qr + rp, 0

(pq + qr + rp) x 0 = (pq x 0) + (qr x 0) + (rp x 0)

= 0

$\huge {\boxed{\green{Thanks}}}$