ONE EIGHTH CONTEST 2025
Achimota School - 68 points
Krobo Girls' Presby SHS - 34 points
Simms SHS - 14 points
ONE EIGHTH CONTEST 2025
Koforidua Sec. Tech. - 38 points
Ofori Panin SHS - 26 points
Frafraha Comm. SHS - 16 points
ONE EIGHTH CONTEST 2025
Keta SHTS - 56 points
Chemu SHTS - 25 points
Accra High School - 9 points
ROUND 1
PREAMBLE
Find the constants P and Q given that:
FIRST QUESTION
Given that the polynomial $x^{3}+px^{2}+qx+6$ is divisible by $x-2$ and $x-1$, find p and q.
SOLUTION
General Formula:
For a cubic polynomial $x^3+px^2+qx+c$ with known roots $x_1$ and $x_2$:
1. Product of roots ($P$) = $x_1 \times x_2$
2. Sum of roots ($S$) = $x_1 + x_2$
3. Find the third hidden root value ($H$) = $\frac{c}{P}$
Then, the coefficient formulas are:
$p = -(S - H)$
$q = P - (S \times H)$
Application to Question:
Given roots $x_1 = 2$ and $x_2 = 1$, and constant $c = 6$:
Product ($P$) = $2 \times 1 = 2$
Sum ($S$) = $2 + 1 = 3$
Hidden value ($H$) = $\frac{6}{2} = 3$
Calculating p:
$p = -(S - H)$
$p = -(3 - 3)$
$p = 0$
Calculating q:
$q = P - (S \times H)$
$q = 2 - (3 \times 3)$
$q = 2 - 9$
$q = -7$
FINAL ANSWER: $p=0$, $q=-7$
SECOND QUESTION
Given that the polynomial $x^{3}+px^{2}+qx-8$ is divisible by $x+4$ and $x-1$, find p and q.
SOLUTION
Given roots $x_1 = -4$ and $x_2 = 1$, and constant $c = -8$:
Product ($P$) = $-4 \times 1 = -4$
Sum ($S$) = $-4 + 1 = -3$
Hidden value ($H$) = $\frac{-8}{-4} = 2$
Calculating p:
$p = -(S - H)$
$p = -(-3 - 2)$
$p = -(-5)$
$p = 5$
Calculating q:
$q = P - (S \times H)$
$q = -4 - (-3 \times 2)$
$q = -4 - (-6)$
$q = 2$
FINAL ANSWER: $p=5$, $q=2$
THIRD QUESTION
Given that the polynomial $x^{3}+px^{2}+qx+12$ is divisible by $x+3$ and $x-2$, find p and q.
SOLUTION
Given roots $x_1 = -3$ and $x_2 = 2$, and constant $c = 12$:
Product ($P$) = $-3 \times 2 = -6$
Sum ($S$) = $-3 + 2 = -1$
Hidden value ($H$) = $\frac{12}{-6} = -2$
Calculating p:
$p = -(S - H)$
$p = -(-1 - (-2))$
$p = -(-1 + 2)$
$p = -1$
Calculating q:
$q = P - (S \times H)$
$q = -6 - (-1 \times -2)$
$q = -6 - 2$
$q = -8$
FINAL ANSWER: $p=-1$, $q=-8$
PRACTICE QUESTIONS
1. Given that the polynomial $x^{3}+px^{2}+qx+6$ is divisible by $x-2$ and $x-1$, find $p$ and $q$.
ANSWER: $p=0$, $q=-7$
SOLUTION 1
The polynomial has roots $2$, $1$ and a third root $h$.
The product of the roots is $-c=-6$:
$2\times1\times h=-6$
$h=-3$
The sum of the roots is $-p$, and the sum of the products in pairs is $q$:
$p=-(2+1-3)=0$
$q=2(1)+2(-3)+1(-3)=-7$
SOLUTION 2
By the factor theorem, $f(2)=0$ and $f(1)=0$:
$8+4p+2q+6=0\Rightarrow 4p+2q=-14$
$1+1p+q+6=0\Rightarrow p+q=-7$
Eliminate $q$: multiply the first equation by $1$ and the second by $2$, then subtract:
$2p=0$
$p=0$
Substitute into the first equation:
$2q=-14$
$q=-7$
2. Given that the polynomial $x^{3}+px^{2}+qx-8$ is divisible by $x+4$ and $x-1$, find $p$ and $q$.
ANSWER: $p=5$, $q=2$
SOLUTION 1
The polynomial has roots $-4$, $1$ and a third root $h$.
The product of the roots is $-c=8$:
$-4\times1\times h=8$
$h=-2$
The sum of the roots is $-p$, and the sum of the products in pairs is $q$:
$p=-(-4+1-2)=5$
$q=-4(1)+-4(-2)+1(-2)=2$
SOLUTION 2
By the factor theorem, $f(-4)=0$ and $f(1)=0$:
$-64+16p-4q-8=0\Rightarrow 16p-4q=72$
$1+1p+q-8=0\Rightarrow p+q=7$
Eliminate $q$: multiply the first equation by $1$ and the second by $-4$, then subtract:
$20p=100$
$p=5$
Substitute into the first equation:
$-4q=-8$
$q=2$
3. Given that the polynomial $x^{3}+px^{2}+qx+6$ is divisible by $x+3$ and $x-2$, find $p$ and $q$.
ANSWER: $p=0$, $q=-7$
SOLUTION 1
The polynomial has roots $-3$, $2$ and a third root $h$.
The product of the roots is $-c=-6$:
$-3\times2\times h=-6$
$h=1$
The sum of the roots is $-p$, and the sum of the products in pairs is $q$:
$p=-(-3+2+1)=0$
$q=-3(2)+-3(1)+2(1)=-7$
SOLUTION 2
By the factor theorem, $f(-3)=0$ and $f(2)=0$:
$-27+9p-3q+6=0\Rightarrow 9p-3q=21$
$8+4p+2q+6=0\Rightarrow 4p+2q=-14$
Eliminate $q$: multiply the first equation by $2$ and the second by $-3$, then subtract:
$30p=0$
$p=0$
Substitute into the first equation:
$-3q=21$
$q=-7$
4. Given that the polynomial $x^{3}+px^{2}+qx+6$ is divisible by $x-3$ and $x+1$, find $p$ and $q$.
ANSWER: $p=-4$, $q=1$
SOLUTION 1
The polynomial has roots $3$, $-1$ and a third root $h$.
The product of the roots is $-c=-6$:
$3\times(-1)\times h=-6$
$h=2$
The sum of the roots is $-p$, and the sum of the products in pairs is $q$:
$p=-(3-1+2)=-4$
$q=3(-1)+3(2)+(-1)(2)=1$
SOLUTION 2
By the factor theorem, $f(3)=0$ and $f(-1)=0$:
$27+9p+3q+6=0\Rightarrow 9p+3q=-33$
$-1+1p-q+6=0\Rightarrow p-q=-5$
Eliminate $q$: multiply the first equation by $-1$ and the second by $3$, then subtract:
$-12p=48$
$p=-4$
Substitute into the first equation:
$3q=3$
$q=1$
5. Given that the polynomial $x^{3}+px^{2}+qx+12$ is divisible by $x-2$ and $x+2$, find $p$ and $q$.
ANSWER: $p=-3$, $q=-4$
SOLUTION 1
The polynomial has roots $2$, $-2$ and a third root $h$.
The product of the roots is $-c=-12$:
$2\times(-2)\times h=-12$
$h=3$
The sum of the roots is $-p$, and the sum of the products in pairs is $q$:
$p=-(2-2+3)=-3$
$q=2(-2)+2(3)+(-2)(3)=-4$
SOLUTION 2
By the factor theorem, $f(2)=0$ and $f(-2)=0$:
$8+4p+2q+12=0\Rightarrow 4p+2q=-20$
$-8+4p-2q+12=0\Rightarrow 4p-2q=-4$
Eliminate $q$: multiply the first equation by $-2$ and the second by $2$, then subtract:
$-16p=48$
$p=-3$
Substitute into the first equation:
$2q=-8$
$q=-4$
6. Given that the polynomial $x^{3}+px^{2}+qx-8$ is divisible by $x+1$ and $x+2$, find $p$ and $q$.
ANSWER: $p=-1$, $q=-10$
SOLUTION 1
The polynomial has roots $-1$, $-2$ and a third root $h$.
The product of the roots is $-c=8$:
$-1\times(-2)\times h=8$
$h=4$
The sum of the roots is $-p$, and the sum of the products in pairs is $q$:
$p=-(-1-2+4)=-1$
$q=-1(-2)+-1(4)+(-2)(4)=-10$
SOLUTION 2
By the factor theorem, $f(-1)=0$ and $f(-2)=0$:
$-1+1p-q-8=0\Rightarrow p-q=9$
$-8+4p-2q-8=0\Rightarrow 4p-2q=16$
Eliminate $q$: multiply the first equation by $-2$ and the second by $-1$, then subtract:
$2p=-2$
$p=-1$
Substitute into the first equation:
$-1q=10$
$q=-10$