Question

Write the number of zeros of the polynomial p(X) X cube +2xsquare + X+6

Answer 1

Answer:

${x}^{3} + 2 {x}^{2} + x + 6 = 0 \\ \\ x( {x}^{2} + 2x + 1) + 6 = 0 \\ \\ x(x + 1) {}^{2} + 6 = 0$

Graph

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

Peak values at

$\frac{dy}{dx} = 0 \\ \\ (x + 1) {}^{2} + 2x(x + 1) = 0 \\ \\ (x + 1)(3x + 1) = 0 \\ \\ x = - 1 \: \: \: \: and \: \: \: \: \frac{ - 1}{ \: \: 3}$

At -1 , value = 0

At -1/3 , value = - 4/27

We need graph of

$x(x + 1) {}^{2} + 6 = y + 6$

So Peak will be at

$0 + 6 = 6 \\and \\ - \frac{4}{27} + 6 = \frac{158}{27}$

Observe Graph

So there will be

1 Real Root