Solve- 5z-22 - 40z-5z - 2-3 Marks-

Given, (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[1 Mark] Factorizing, 25z^2+20z+4 ⇒(5z)^2 + 2×5z × ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard VIII

Mathematics

Factors of a³ - b³

Solve: 5z-22 ...

Question

Solve: $\frac{({(5 z - 2)}^{2} + 40 z)}{(5 z + 2)}$ $[3 M a r k s]$

Open in App

Solution

$G i v e n, (5 z - 2)^{2} + 40 z$
$[We know, (a - b)^{2} = a^{2} - 2 a b + b^{2}]$
$\Rightarrow 25 z^{2} - 20 z + 4 + 40 z$
$\Rightarrow 25 z^{2} + 20 z + 4$ $[1 M a r k]$

$F a c t o r i z i n g, 25 z^{2} + 20 z + 4$
$\Rightarrow (5 z)^{2} + 2 \times 5 z \times 2 + 2^{2}$
$[We know, (a + b)^{2} = a^{2} + 2 a b + b^{2}]$
$\Rightarrow 25 z^{2} + 20 z + 4 = (5 z + 2)^{2}$
$∴ (5 z - 2)^{2} + 40 z = (5 z + 2)^{2}$ $[1 M a r k]$

$So, \frac{({(5 z - 2)}^{2} + 40 z)}{(5 z + 2)} = \frac{(5 z + 2)^{2}}{(5 z + 2)}$
$= 5 z + 2$ $[1 M a r k]$

Suggest Corrections

Solve: ((5z−2)2+40z)(5z+2) [3 Marks]

Given,(5z−2)2+40z [We know, (a−b)2=a2−2ab+b2] ⇒25z2−20z+4+40z ⇒25z2+20z+4 [1 Mark] Factorizing, 25z2+20z+4 ⇒(5z)2+2×5z ×2+22 [We know, (a+b)2=a2+2ab+b2] ⇒25z2+20z+4=(5z+2)2 ∴(5z−2)2+40z=(5z+2)2 [1 Mark] So, ((5z−2)2+40z)(5z+2)=(5z+2)2(5z+2) =5z+2 [1 Mark]

Solve: $\frac{({(5 z - 2)}^{2} + 40 z)}{(5 z + 2)}$ $[3 M a r k s]$