Question

finde the zero of the polynomial of 4u square+8a​

Answer 1

Answer:

$\huge{\boxed{\boxed{\sf{u=-2\:and\:0}}}}$

Step-by-step explanation:

$\huge{\boxed{\boxed{\underline{\sf{SOLUTION-}}}}}$

>>Correction in question:

$4u^{2}+8a$

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$4 {u}^{2} + 8u = 0 \\ = )4u(u + 2) = 0 \\ = )4u = 0 \\ = )u = 0 - - - - - 1st \: zeroes\\ = ) u + 2 = 0 \\ = ) u = - 2 - - - - - 2nd \: zeroes$

$\huge{\boxed{\boxed{\sf{u=-2\:and\:0}}}}$

$\huge{\boxed{\boxed{\underline{\sf{VERIFICATION}}}}}$

$4 {u}^{2} + 8u \\ first \: putting \: the \: value \: of \: u = 0 \\ = )4( {0})^{2} + 8 \times 0 \\ = )0 + 0 \\ = ) 0 \\ again \: put \: the \: value \: of \: u = - 2 \\ 4 {u}^{2} + 8u \\ = )4( { - 2})^{2} + 8 \times - 2 \\ = ) 4 \times 4 - 16 \\ = )16 - 16 \\ = )0$

$\huge{\boxed{\boxed{\underline{\sf{VERIFIED}}}}}$

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