Question

Find the product of 8s2(t2-2st) and verify the result when S=1 and t=5

Answer 1

Answer:

$\red { Product \: of \: 8s^{2}(t^{2}-2st)}\green{=8s^{2}t^{2} - 16s^{3}t}$

Step-by-step explanation:

$\red { Product \: of \: 8s^{2}(t^{2}-2st)}$

$= 8s^{2} \times t^{2} - 8s^{2} \times 2st$

$= 8s^{2}t^{2} - 16s^{3}t$

$\red { Product \: of \: 8s^{2}(t^{2}-2st)}\green{=8s^{2}t^{2} - 16s^{3}t}\:---(1)$

Verification:

$Substitute \: s = 1 \:and \: t=5\:in \: equation \:(1)$

$\blue {LHS} = 8 \times (1)^{2} (5^{2}-2\times 1 \times 5)\\=8(25 - 10)\\= 8 \times 15 \\= 120$

$\blue {RHS} = {8s^{2}t^{2} - 16s^{3}t}\\= 8\times 1^{2}\times 5^{2} - 16 \times 1^{3}\times 5\\= 8\times 25 - 16 \times 5 \\= 200 - 80 \\= 120$

$\blue { RHS = LHS }$

Answer 2

Answer:

Productof8s

2

(t

2

−2st)=8s

2

t

2

−16s

3

t

Step-by-step explanation:

\red { Product \: of \: 8s^{2}(t^{2}-2st)}Productof8s

2

(t

2

−2st)

= 8s^{2} \times t^{2} - 8s^{2} \times 2st=8s

2

×t

2

−8s

2

×2st

= 8s^{2}t^{2} - 16s^{3}t=8s

2

t

2

−16s

3

t

\red { Product \: of \: 8s^{2}(t^{2}-2st)}\green{=8s^{2}t^{2} - 16s^{3}t}\:---(1)Productof8s

2

(t

2

−2st)=8s

2

t

2

−16s

3

t−−−(1)

Verification:

Substitute \: s = 1 \:and \: t=5\:in \: equation \:(1)Substitutes=1andt=5inequation(1)

\begin{gathered}\blue {LHS} = 8 \times (1)^{2} (5^{2}-2\times 1 \times 5)\\=8(25 - 10)\\= 8 \times 15 \\= 120\end{gathered}

LHS=8×(1)

2

(5

2

−2×1×5)

=8(25−10)

=8×15

=120

\begin{gathered} \blue {RHS} = {8s^{2}t^{2} - 16s^{3}t}\\= 8\times 1^{2}\times 5^{2} - 16 \times 1^{3}\times 5\\= 8\times 25 - 16 \times 5 \\= 200 - 80 \\= 120\end{gathered}

RHS=8s

2

t

2

−16s

3

t

=8×1

2

×5

2

−16×1

3

×5

=8×25−16×5

=200−80

=120