Question

solve all zeros of the polynomial function.
$f(x) = x^4 + 5x^3 - 7x^2 - 29x + 30$

Explanation:

Step1: Use Rational Root Theorem

Possible rational zeros: $\pm1, \pm2, \pm3, \pm5, \pm6, \pm9, \pm10, \pm15, \pm18, \pm30, \pm45, \pm90$

Step2: Test $x=1$

$f(1)=1^4 + 5(1)^3 -7(1)^2 -25(1)+90=1+5-7-25+90=64
eq0$

Step3: Test $x=-5$

$f(-5)=(-5)^4 + 5(-5)^3 -7(-5)^2 -25(-5)+90=625-625-175+125+90=40
eq0$

Step4: Test $x=3$

$f(3)=3^4 + 5(3)^3 -7(3)^2 -25(3)+90=81+135-63-75+90=168
eq0$

Step5: Test $x=-3$

$f(-3)=(-3)^4 + 5(-3)^3 -7(-3)^2 -25(-3)+90=81-135-63+75+90=48
eq0$

Step6: Test $x=2$

$f(2)=2^4 + 5(2)^3 -7(2)^2 -25(2)+90=16+40-28-50+90=68
eq0$

Step7: Test $x=-2$

$f(-2)=(-2)^4 + 5(-2)^3 -7(-2)^2 -25(-2)+90=16-40-28+50+90=88
eq0$

Step8: Test $x=5$

$f(5)=5^4 + 5(5)^3 -7(5)^2 -25(5)+90=625+625-175-125+90=1040
eq0$

Step9: Test $x=-4$

$f(-4)=(-4)^4 + 5(-4)^3 -7(-4)^2 -25(-4)+90=256-320-112+100+90=14
eq0$

Step10: Factor by grouping

$x^4+5x^3-7x^2-25x+90 = x^3(x+5) -7x(x+5) + 18(x+5)=(x+5)(x^3-7x+18)$

Step11: Factor cubic polynomial

Test $x=-3$ for $x^3-7x+18$: $(-3)^3-7(-3)+18=-27+21+18=12
eq0$
Test $x=-2$ for $x^3-7x+18$: $(-2)^3-7(-2)+18=-8+14+18=24
eq0$
Test $x=3$ for $x^3-7x+18$: $3^3-7(3)+18=27-21+18=24
eq0$
Test $x=2$ for $x^3-7x+18$: $2^3-7(2)+18=8-14+18=12
eq0$
Use quadratic formula on $x^3-7x+18=(x+3)(x^2-3x+6)$

Step12: Find real zeros

$x+5=0 \implies x=-5$; $x+3=0 \implies x=-3$
Quadratic $x^2-3x+6$ has discriminant $\Delta=(-3)^2-4(1)(6)=9-24=-15<0$, no real zeros.

Answer:

-5, -3