Question

Solve (x3 +x²+x+1) (x+2) /(x²+1) is less than Or equal to 0.​

Answer 1

Answer:

Apply remainder theorem

x+1=0

x=−1

Put the value of x=−1 in all equations.

(i) x

3

+x

2

+x+1=(−1)

3

+(−1)

2

+(−1)+1=−1+1−1+1=0

Then x+1 is the factor of equation

(ii) x

4

+x

3

+x

2

+x+1=(−1)

4

+(−1)

3

+(−1)

2

+(−1)+1=1−1+1−1+1=1

This is not zero.Then x+1 is not the factor of equation

(iii) x

4

+3x

3

+3x

2

+x+1=(−1)

4

+3(−1)

3

+3(−1)

2

+(−1)+1=1

This is not zero.Then x+1 is not the factor of equation

(iv)x

3

−x

2

−(2+

2

)x+

2

=(−1)

3

−(−1)

2

−(2+

2

)(−1)+

2

=−1−1+2−

2

+

2

=0

( make me Barinlist and I'm army and blink )

Answer 2

$\large\underline{\sf{Solution-}}$

Given inequality is

$\rm :\longmapsto\:\dfrac{( {x}^{3} + {x}^{2} + x + 1)(x + 2)}{ {x}^{2} + 1} \leqslant 0$

As,

$\purple{\rm :\longmapsto\: {x}^{2} + 1 > 0 \: \: \forall \: x \: \in \: R}$

So,

$\rm :\longmapsto\:( {x}^{3} + {x}^{2} + x + 1)(x + 2) \leqslant 0$

$\rm :\longmapsto\:[ {x}^{2}(x + 1) + 1(x + 1)](x + 2) \leqslant 0$

$\rm :\longmapsto\:(x + 1)( {x}^{2} + 1)(x + 2) \leqslant 0$

As,

$\purple{\rm :\longmapsto\: {x}^{2} + 1 > 0 \: \: \forall \: x \: \in \: R}$

So,

$\rm :\longmapsto\:(x + 1)(x + 2) \leqslant 0$

$\rm\implies \: - 2 \leqslant x \leqslant - 1$

$\bf\implies \:x \: \in \: [ - 2, \: - 1]$

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Concept Used

If a and b are two real positive numbers such that a < b, then

$\boxed{\tt{ (x - a)(x - b) < 0 \: \rm\implies \:a < x < b \: }}$

$\boxed{\tt{ (x - a)(x - b) \leqslant 0 \: \rm\implies \:a \leqslant x \leqslant b \: }}$

$\boxed{\tt{ (x - a)(x - b) > 0 \: \rm\implies \:x < a \: or \: x > b \: }}$

$\boxed{\tt{ (x - a)(x - b) \geqslant 0 \: \rm\implies \:x \leqslant a \: or \: x \geqslant b \: }}$

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More to Know

$\purple{\rm :\longmapsto\: |x| < y\rm\implies \: - y < x < y \: }$

$\purple{\rm :\longmapsto\: |x| \leqslant y\rm\implies \: - y \leqslant x \leqslant y \: }$

$\purple{\rm :\longmapsto\: |x - z| < y\rm\implies \:z - y < x < z + y \: }$

$\purple{\rm :\longmapsto\: |x - z| \leqslant y\rm\implies \:z - y \leqslant x \leqslant z + y \: }$