Question

How to take out the zero of the polynomial:
$l(x)=x^{2} +2x+1$

Answer 1

Answer:

2x + 1, for 0 £ x £ 1 32x – 29, for 1 £ x £ 2 (ii) In0£x£1,x0=0,x1=1;y0=1,y1=3; y¢0 = 1, y¢ 1 ... x; l¢0(x) = – 1; l¢1(x) = 1 u0(x)= 1 – 2x × (– 1 ) = 1 + 2x; u1(x)=1–2(x–1)×1 ...

Answer 2

Answer:

$\huge{\fbox{\fbox{\pink{HELLO}}}}$

$l(x)=x^{2} +2x+1$

x^{2} + x + x + 1

x ( x + 1 ) +1(x +1 )

(x +1)(x+1)

zeros are -1,-1

hope it helps

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