What are all the possible rational zeros for $f (x) = x^{3} + 13 x^{2} + 38 x + 24$ $f(x)=x^3+13x^2+38x+24$ and how do you find all zeros?

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Answer 1 · 2017-09-16T12:33:11

Zeros are $- 3, - 9.123, - 0.877$ $-3 , -9.123, -0.877$

Possible rational zeros are found as under ,

Factors of constant term $24$ $24$ are $\pm \frac{1, 2, 3, 4, 6, 8, 12, 24}{}$ $+-( 1,2,3,4,6,8,12,24)/$

factor of leading coefficient $1$ $1$ is $\pm 1$ $+-1$ , so possible

rational zeros are $\pm (1, 2, 3, 4, 6, 8, 12, 24)$ $+-( 1,2,3,4,6,8,12,24)$ . By trying long divisions

$- 3$ $-3$ suits only . $f (x) = x^{3} + 13 x^{2} + 38 x + 24$ $f(x) = x^3 +13x^2 +38x+24$ or

$f (x) = x^{3} + 3 x^{2} + 10 x^{2} + 30 x + 8 x + 24$ $f(x) = x^3 +3x^2 +10x^2+30x+8x+24$ or

$f (x) = x^{2} (x + 3) + 10 x (x + 3) + 8 (x + 3)$ $f(x) = x^2(x+3)+ 10x(x+3)+8(x+3)$ or

$f (x) = (x + 3) (x^{2} + 10 x + 8)$ $f(x) = (x+3)(x^2+10x +8)$ , for $(x^{2} + 10 x + 8)$ $(x^2+10x +8)$

$x = \frac{- 10 \pm \sqrt{100 - 32}}{2}$ $x= (-10 +- sqrt( 100 -32))/2$ or

$x ~~ -9.123 , x ~~ -0.877 (3dp) :.$

$f(x) = x^3 +13x^2 +38x+24 = (x+3)(x+9.123)(x+0.877)$

Zeros are $-3 , -9.123, -0.877$ [Ans]

What are all the possible rational zeros for f(x)=x^3+13x^2+38x+24f(x)=x3+13x2+38x+24f(x)=x^3+13x^2+38x+24 and how do you find all zeros?