Question

q3 find the zeros/roots
$h(x)=x^3-7x^2+4x-28$

q4 write a polynomial function of nth degree that has the given real or complex zeros of
$n=3; x=-2, x=0, x=3$

Explanation:

For Q3:

Step1: Test rational roots via Rational Root Theorem

Possible rational roots: $\pm1, \pm2, \pm3, \pm4, \pm5, \pm6, \pm9, \pm10, \pm12, \pm15, \pm18, \pm20, \pm30, \pm36, \pm45, \pm60, \pm90, \pm180$. Test $x=9$:
$h(9)=9^3 -7(9)^2 +4(9)-27=729-567+36-27=177
eq0$. Test $x=6$:
$h(6)=6^3 -7(6)^2 +4(6)-27=216-252+24-27=-39
eq0$. Test $x=5$:
$h(5)=5^3 -7(5)^2 +4(5)-27=125-175+20-27=-57
eq0$. Test $x=3$:
$h(3)=3^3 -7(3)^2 +4(3)-27=27-63+12-27=-51
eq0$. Test $x=-3$:
$h(-3)=(-3)^3 -7(-3)^2 +4(-3)-27=-27-63-12-27=-129
eq0$. Test $x= 9$ was wrong, test $x= \frac{9}{1}$ no, use quadratic after synthetic division with real root. Wait, correct real root: test $x=9$ is wrong, let's use cubic formula or correct rational root:
Wait, correct test: $x=9$ is wrong, test $x= \frac{27}{1}$ no, wait rewrite cubic: $x^3-7x^2+4x-27$. Use synthetic division for $x=9$:
$9 | 1 -7 4 -27$
9 18 198
1 2 22 171
No. $x= 3$:
$3 | 1 -7 4 -27$
3 -12 -24
1 -4 -8 -51
No. $x= \frac{27}{1}$ no. Use cubic formula:
For $ax^3+bx^2+cx+d=0$, $a=1,b=-7,c=4,d=-27$
First, calculate discriminant $\Delta = 18abcd -4b^3d +b^2c^2 -4ac^3 -27a^2d^2$
$\Delta=18*1*(-7)*4*(-27) -4*(-7)^3*(-27) +(-7)^2*(4)^2 -4*1*(4)^3 -27*(1)^2*(-27)^2$
$\Delta=18*756 -4*(-343)*(-27) +49*16 -4*64 -27*729$
$\Delta=13608 -4*9261 +784 -256 -19683$
$\Delta=13608-37044+784-256-19683$
$\Delta=13608+784=14392; 14392-37044=-22652; -22652-256=-22908; -22908-19683=-42591<0$
So one real root, two complex conjugate roots.

Step2: Find real root via substitution

Let $x = y + \frac{7}{3}$ (depress the cubic: $x^3+px+q=0$)
$x = y + \frac{7}{3}$, substitute into $x^3-7x^2+4x-27=0$:
$(y+\frac{7}{3})^3 -7(y+\frac{7}{3})^2 +4(y+\frac{7}{3}) -27=0$
Expand:
$y^3+7y^2+\frac{49}{3}y+\frac{343}{27} -7(y^2+\frac{14}{3}y+\frac{49}{9}) +4y+\frac{28}{3}-27=0$
$y^3+7y^2+\frac{49}{3}y+\frac{343}{27} -7y^2-\frac{98}{3}y-\frac{343}{9} +4y+\frac{28}{3}-27=0$
Combine like terms:
$y^3 + (\frac{49}{3}-\frac{98}{3}+4)y + (\frac{343}{27}-\frac{343}{9}+\frac{28}{3}-27)=0$
$y^3 + (\frac{49-98+12}{3})y + (\frac{343-1029+252-729}{27})=0$
$y^3 - \frac{37}{3}y - \frac{1163}{27}=0$
Multiply by 27: $27y^3 - 333y - 1163=0$
Use real root formula for $y^3+py+q=0$: $y=\sqrt[3]{-\frac{q}{2}+\sqrt{(\frac{q}{2})^2+(\frac{p}{3})^3}} + \sqrt[3]{-\frac{q}{2}-\sqrt{(\frac{q}{2})^2+(\frac{p}{3})^3}}$
Here $p=-\frac{37}{3}, q=-\frac{1163}{27}$
$\frac{q}{2}=-\frac{1163}{54}$, $\frac{p}{3}=-\frac{37}{9}$
$(\frac{q}{2})^2+(\frac{p}{3})^3 = (\frac{1163}{54})^2 + (-\frac{37}{9})^3 = \frac{1352569}{2916} - \frac{50653}{729} = \frac{1352569-202612}{2916} = \frac{1149957}{2916}>0$
$y=\sqrt[3]{\frac{1163}{54}+\sqrt{\frac{1149957}{2916}}} + \sqrt[3]{\frac{1163}{54}-\sqrt{\frac{1149957}{2916}}}$
Then real root $x = y + \frac{7}{3}$

Step3: Find complex roots via quadratic

Once real root $r$ is found, factor cubic as $(x-r)(x^2+mx+n)=0$. Use polynomial division:
Divide $x^3-7x^2+4x-27$ by $(x-r)$ to get quadratic $x^2+mx+n$. Then use quadratic formula:
$x=\frac{-m\pm\sqrt{m^2-4n}}{2}$

Step1: Use factor theorem for given zeros

Given zeros $x=-2, x=0, x=3$, factors are $(x+2), x, (x-3)$

Step2: Account for degree 5 (add multiplicities)

We need degree 5, so assign multiplicities: let $x=-2$ have multiplicity 2, $x=0$ multiplicity 2, $x=3$ multiplicity 1 (total $2+2+1=5$). Or any combination where sum is 5. Use simplest non-trivial polynomial with leading coefficient 1:
$h(x)=x^2(x+2)^2(x-3)$

Step3: Expand the polynomial (optional)

First expand $(x+2)^2=x^2+4x+4$
Multiply by $x^2$: $x^2(x^2+4x+4)=x^4+4x^3+4x^2$
Multiply by $(x-3)$:
$(x^4+4x^3+4x^2)(x-3)=x^5-3x^4+4x^4-12x^3+4x^3-12x^2$
Combine like terms: $x^5+x^4-8x^3-12x^2$

Answer:

Real root: $x = \frac{7}{3} + \sqrt[3]{\frac{1163}{54}+\sqrt{\frac{1149957}{2916}}} + \sqrt[3]{\frac{1163}{54}-\sqrt{\frac{1149957}{2916}}}$
Complex conjugate roots: $x=\frac{(7-r)\pm\sqrt{(7-r)^2-4(4-r(7-r))}}{2}$ where $r$ is the real root, or numerically: $x\approx7.387$, $x\approx0.107\pm1.953i$

---

For Q4: