Question

Multiply (pq+4) (p + r*r) (q + 8)


Plz plz ans fast!!!!!!

Answer 1

Answer:

p²q² + pq²r² + 4pq + 4qr² + 8p²q + 8pqr² + 32p + 32r²

Step-by-step explanation:

Given : (pq + 4)(p + r*r)(q + 8)

→ (pq + 4)(p + r²)(q + 8)

→ (pq + 4)(p + r²)(q + 8)

→ [ (pq + 4)(p + r²) ] (q + 8)

→ [ pq(p + r²) + 4(p + r²) ] (q + 8)

→ [ (pq)(p) + (pq)(r²) + (4)(p) + (4)(r²) ] (q + 8)

→ [ p²q + pqr² + 4p + 4r² ] (q + 8)

→ q(p²q + pqr² + 4p + 4r²) + 8(p²q + pqr² + 4p + 4r²)

→ (q)(p²q) + (q)(pqr²) + (q)(4p) + (q)(4r²) + (8)(p²q) + (8)(pqr²) + (8)(4p) + (8)(4r²)

→ p²q² + pq²r² + 4pq + 4qr² + 8p²q + 8pqr² + 32p + 32r²

Answer 2

AnsWer :

$\:\bullet$ We can break the question into three parts [i.e. (pq + 4) = I, (p+ r × r) = II, (q + 8) = III].

$\underline{\bigstar\:\textsf{According \: to \: given \: in \: question:}}$

$\normalsize\ : \implies\sf\underbrace{(pq + 4)}_{ \red{part \: I}} \: \: \underbrace{(p + r^2)}_{ \green{part \: 2} } \: \; \underbrace{(q + 8)}_ {\blue{part \: 3} }$

$\rule{100}2$

★Multiply part I & II :

$\normalsize\ : \implies\sf\underbrace{(pq+4)(p + r^2)}$

$\normalsize\ : \implies\sf\ p(pq + 4) + r^2(pq + 4) \\ \\ \normalsize\ : \implies\sf\ p^2q + 4p + pqr^2 + 4r^2$

$\rule{100}2$

★Multiply the product of I & II with III :

$\normalsize\ : \implies\sf\underbrace{(p^2q + 4p + pqr^2 + 4r^2)(q+8)}$

$\normalsize\ : \implies\sf\ q(p^2q + 4p + pqr^2 + 4r^2) + 8(p^2q + 4p + pqr^2 + 4r^2) \\ \\ \normalsize\ : \implies\sf\ p^2q^2 + 4pq + pq^2r^2 + 4qr^2 + 8p^2q + 32p + 8pqr^2 + 32r^2$

$\rule{100}2$

$\therefore\sf\ Your \: Answer:$ $\normalsize{\underline{\boxed{\sf \purple{p^2q^2 + 4pq + pq^2r^2 + 4qr^2 + 8p^2q + 32p + 8pqr^2 + 32r^2}}}}$