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)