WAEC Further Mathematics Past Questions & Answers

(x - 5) is a factor of $x^3 - 4x^2 - 11x + 30$. To find the remaining factors, let's draw out $(x - 5)$ from the parent expression.

$x^3 - 4x^2 - 11x + 30 = x^3 - 5x^2 + x^2 - 5x - 6x + 30$

$= x^2(x - 5) + x(x - 5) - 6(x - 5) = (x - 5)(x^2 + x - 6)$

∴ To find the remaining factors, we factorize $(x2 + x - 6)$

$x^2 + x - 6 = x^2 + 3x - 2x - 6$

$= x(x + 3) - 2(x + 3) = (x + 3)(x - 2)$

∴ The other two factors are $(x + 3) and (x - 2)$

ALTERNATIVELY
$∴ x^2 + x - 6 = (x + 3) and (x - 2)$

$=^6C_4$

$=\frac{6!}{(4!\times2!)}$

$=\frac{6\times5}{2\times1}$

= 15

rth term of a binomial expansion =$^nC_r-1 a^{n-(r-1)}b^{r-1}$

$n = 10,r = 6 \therefore r-1=5$

6th term =$^{10}C_5 1^{10} - 5 (-\frac{2}{3}x)^5$

$=252*1*-\frac{32x^5}{243}=-\frac{896x^5}{27}$

$x^2-$(sum of roots)$x+$(product of roots) = $0$

$4x^2-4x-15=0$

Divide through by 4

$=x^2-x-\frac{15}{4}=0$

$=x^2-x+(-\frac{15}{4})=0$

$=x^2-(1)x+(-\frac{15}{4})=0$

sum of roots =1

= m + (m + 4) = 1
=2m+4=1

=2m=-3

=m=-$\frac{3}{2}$

The equation whose roots are 2m and 2m+8

2m=2×-$\frac{3}{2}=-3$and $2m+8=2×-\frac{3}{2}+8=5$

$=x^2-(-3+5)x+(-3)(5)=0$

$=x^2-2x+(-15)=0$

$∴x^2-2x-15=0$

$p = \begin{bmatrix} x&4\\3&7\end{bmatrix}, Q =\begin{bmatrix} x&3\\1&2x\end{bmatrix}$

$p = \begin{bmatrix} x&4\\3&7\end{bmatrix}=7x-12, Q =\begin{bmatrix} x&3\\1&2x\end{bmatrix}=2x^2-3$

|Q| = |P| + 3 (Given)