The correct option is D 5z + 2Given, (5z 2)^2+40z [We know, (a b)^2=a^2 nbsp;2ab + b^2] ⇒ 25z^2 nbsp;20z + 4 + 40z ⇒ 25z^2 + 20z + 4 Factorizing, 25z^2+ ...

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Byju's Answer

Standard VIII

Mathematics

Expansion of (x-y)^2

.52-22+40 z/5...

Question

$\frac{{{(5 z - 2)}^{2} + 40 z}}{(5 z + 2)} =$

2 z + 5

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(5 z - 2)^{2}

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5 z - 2

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5 z + 2

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Solution

The correct option is D $5 z + 2$
$G i v e n, (5 z - 2)^{2} + 40 z$
$= {(5 z)^{2} - (2 \times 5 z \times 2) + (2)^{2}} + 40 z$ $[We know, (a - b)^{2} = a^{2} - 2 a b + b^{2}]$
$= 25 z^{2} - 20 z + 4 + 40 z$
$= 25 z^{2} + 20 z + 4$
$= (5 z)^{2} + 2 \times 5 z \times 2 + 2^{2}$ [on Factorizing]
$= (5 z + 2)^{2}$ $[We know, a^{2} + 2 a b + b^{2} = (a + b)^{2}]$

$∴ (5 z - 2)^{2} + 40 z = (5 z + 2)^{2}$

$So, \frac{{{(5 z - 2)}^{2} + 40 z}}{(5 z + 2)} = \frac{(5 z + 2)^{2}}{(5 z + 2)} = 5 z + 2$

Suggest Corrections

{(5z−2)2+40z}(5z+2)=

The correct option is D 5z+2Given,(5z−2)2+40z={(5z)2−(2×5z×2)+(2)2}+40z [We know, (a−b)2=a2−2ab+b2] =25z2−20z+4+40z =25z2+20z+4=(5z)2+2×5z ×2+22 [on Factorizing]=(5z+2)2 [We know, a2+2ab+b2=(a+b)2] ∴(5z−2)2+40z=(5z+2)2 So, {(5z−2)2+40z}(5z+2)=(5z+2)2(5z+2)=5z+2

$\frac{{{(5 z - 2)}^{2} + 40 z}}{(5 z + 2)} =$