Factor a real quartic by a conjugate pair · source reference

Question

Do not use a calculator in answering this question.

(a) $f(x)$ is a degree-4 polynomial with real coefficients. Explain whether $f(x)=0$ can have 3 non-real roots and 1 real root. [1]

(b) One root of $2x^4-15x^3+ax^2-63x+b=0$, where $a,b$ are real, is $3-2i$. Find the other roots and $a,b$. [6]

[2024/YIJC/Prelim/I/3]

Source working

Printed working

(a) Not possible. Conjugate root theorem: all-real coefficients require non-real roots in conjugate pairs, so their number cannot be odd.

(b) Method 1: The conjugate $3+2i$ is a root.

$$\begin{aligned}[x-(3-2i)][x-(3+2i)]&=(x-3+2i)(x-3-2i)\\&=(x-3)^2-(2i)^2\\&=x^2-6x+9-(-4)=x^2-6x+13.\end{aligned}$$ $$2x^4-15x^3+ax^2-63x+b=(x^2-6x+13)(2x^2+cx+b/13).$$

Compare $x^3$: $-15=-6(2)+c\Rightarrow c=-3$. Compare $x$: $-63=-6(b/13)+13c=-6b/13+13(-3)$; so $-6b=-312$, $b=52$. Compare $x^2$:

$$a=b/13-6c+2(13)=52/13-6(-3)+26=48.$$

Thus $a=48,b=52$ and

$$(x^2-6x+13)(2x^2-3x+4)=0.$$

First pair $3\pm2i$. Second pair:

$$x=\frac{-(-3)\pm\sqrt{(-3)^2-4(2)(4)}}{2(2)}=\frac{3\pm\sqrt{-23}}4=\frac34\pm\frac{\sqrt{23}}4i.$$

Other roots are $3+2i,(3+i\sqrt{23})/4,(3-i\sqrt{23})/4$.

Method 2: Substitute $3-2i$:

$$2(3-2i)^4-15(3-2i)^3+a(3-2i)^2-63(3-2i)+b=0,$$ $$2(-119-120i)-15(-9-46i)+a(5-12i)-63(3-2i)+b=0,$$ $$(-292+5a+b)+(576-12a)i=0.$$

Imaginary: $576-12a=0\Rightarrow a=48$. Real: $-292+5a+b=0\Rightarrow-292+5(48)+b=0\Rightarrow b=52$. The source repeats the conjugate-factor derivation above, then writes

$$2x^4-15x^3+48x^2-63x+52=(x^2-6x+13)(2x^2+cx+4).$$

Compare $x$: $-63=-6(4)+13c\Rightarrow c=-3$. Then $2x^2-3x+4=0$ and the same quadratic formula gives $(3\pm i\sqrt{23})/4$.

Printed answer key

(b) Other roots $3+2i,3/4+i\sqrt{23}/4,3/4-i\sqrt{23}/4$. The short key does not list $a,b$.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: COMPLEX NUMBERS. Question pages 3; solution pages 14, 15, 16. These are original document page numbers, retained as text metadata.