How do you solve $x^{3} + 7 x^{2} + 10 x \geq 0$ $x^3+7x^2+10x>=0$ ?

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How do you solve $x^{3} + 7 x^{2} + 10 x \geq 0$ $x^3+7x^2+10x>=0$ ?

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Answer 1 · 2016-10-01T09:31:04

Solution is $- 5 \geq x \geq - 2$ $-5 >= x>=-2$ or $x \geq 0$ $x >= 0$

Explanation:

Let us first factorize $x^{3} + 7 x^{2} + 10 x$ $x^3+7x^2+10x$ .

$x^{3} + 7 x^{2} + 10 x = x (x^{2} + 7 x + 10) = x (x^{2} + 2 x + 5 x + 10)$ $x^3+7x^2+10x=x(x^2+7x+10)=x(x^2+2x+5x+10)$

= $x (x (x + 2) + 5 (x + 2) = x (x + 2) (x + 5)$ $x(x(x+2)+5(x+2)=x(x+2)(x+5)$

Hence we have to solve the inequality $x^{3} + 7 x^{2} + 10 x > 0$ $x^3+7x^2+10x>0$ or $(x + 5) (x + 2) x \geq 0$ $(x+5)(x+2)x>=0$

From this we know that for the product $(x + 5) (x + 2) x \geq 0$ $(x+5)(x+2)x>=0$ , signs of binomials $(x + 5)$ $(x+5)$ , $(x + 2)$ $(x+2)$ and $x$ $x$ will change around the values $- 5$ $-5$ . $- 2$ $-2$ and $0$ $0$ respectively. In sign chart we divide the real number line around these values, i.e. below $- 5$ $-5$ , between $- 5$ $-5$ and $- 2$ $-2$ , between $- 2$ $-2$ and $0$ $0$ and above $0$ $0$ and see how the sign of $(x + 5) (x + 2) x$ $(x+5)(x+2)x$ changes.

Sign Chart

$X X X X X X X X X X X - 5 X X X X X - 2 X X X X X 0$ $color(white)(XXXXXXXXXXX)-5color(white)(XXXXX)-2color(white)(XXXXX)0$

$(x + 5) X X X X - i v e X X X X + i v e X X + i v e X X X + i v e$ $(x+5)color(white)(XXXX)-ive color(white)(XXXX)+ive color(white)(XX)+ive color(white)(XXX)+ive$

$(x + 2) X X X X - i v e X X X X - i v e X X - i v e X X X + i v e$ $(x+2)color(white)(XXXX)-ive color(white)(XXXX)-ive color(white)(XX)-ive color(white)(XXX)+ive$

$x X X X X X X X - i v e X X X X - i v e X X + i v e X X X + i v e$ $xcolor(white)(XXXXXXX)-ive color(white)(XXXX)-ive color(white)(XX)+ive color(white)(XXX)+ive$

$(x + 5) (x + 2) x$ $(x+5)(x+2)x$
$X X X X X X X X - i v e X X X X + i v e X X - i v e X X X + i v e$ $color(white)(XXXXXXXX)-ive color(white)(XXXX)+ive color(white)(XX)-ive color(white)(XXX)+ive$

It is observed that $(x + 5) (x + 2) x \geq 0$ $(x+5)(x+2)x>= 0$ when either $- 5 \geq x \geq - 2$ $-5 >= x>=-2$ or $x \geq 0$ $x >= 0$ , which is the solution for the inequality.

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How do you solve $x^{3} + 7 x^{2} + 10 x \geq 0$ $x^3+7x^2+10x>=0$ ?

1 Answer

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