If \((x - 5)\) is a factor of \(x^3 - 4x^2 - 11x + 30\), find the remaining factors.
\((x + 3) and (x - 2)\)
\((x - 3) and (x + 2)\)
\((x - 3) and (x - 2)\)
\((x + 3) and (x + 2)\)
Correct answer is A
(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
\(∴ x^2 + x - 6 = (x + 3) and (x - 2)\)
In how many ways can four Mathematicians be selected from six ?
90
60
15
360
Correct answer is C
\(=^6C_4\)
\(=\frac{6!}{(4!\times2!)}\)
\(=\frac{6\times5}{2\times1}\)
= 15
\(-\frac{896x^6}{9}\)
\(-\frac{896x^5}{9}\)
\(-\frac{896x^5}{27}\)
\(-\frac{896x^6}{27}\)
Correct answer is C
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 =\(^{10}C_5 1^{10} - 5 (-\frac{2}{3}x)^5\)
\(=252*1*-\frac{32x^5}{243}=-\frac{896x^5}{27}\)
\(x^2+8x-15=0\)
\(x^2-2x-15=0\)
\(x^2-8x-15=0\)
\(x^2+2x+15=0\)
Correct answer is B
\(x^2-\)(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 \(2m+8=2×-\frac{3}{2}+8=5\)
\(=x^2-(-3+5)x+(-3)(5)=0\)
\(=x^2-2x+(-15)=0\)
\(∴x^2-2x-15=0\)
\(-2,\frac{3}{2}\)
\(2,\frac{3}{2}\)
\(-2,-\frac{3}{2}\)
\(2-,\frac{3}{2}\)
Correct answer is B
\(p = \begin{bmatrix} x&4\\3&7\end{bmatrix}, Q =\begin{bmatrix} x&3\\1&2x\end{bmatrix}\)
\(p = \begin{bmatrix} x&4\\3&7\end{bmatrix}=7x-12, Q =\begin{bmatrix} x&3\\1&2x\end{bmatrix}=2x^2-3\)
|Q| = |P| + 3 (Given)
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