5 Marks Questions

Question 15 Marks

Salma and Abid factorise the algebraic expression $p^4+9 p^2+18$
Salma
$
\begin{aligned}
p^4+9 p^2+18 & =p^4+6 \rho^2+3 p^2+18 \\
& =p^2\left(p^2+6\right)+3\left(p^2+6\right) \\
& =\left(p^2+3\right)\left(p^2+6\right)
\end{aligned}
$
Abid
$
\begin{aligned}
p^4+9 p^2+18 & =p^4+6 p+3 p+18 \\
& =p\left(p^3+6\right)+3(p+6) \\
& =\left(p^3+6\right)(p+6)(p+3)
\end{aligned}
$
Who is correct? Give a reason to justify your answer.

Answer

The algebraic expression is given as $p^4+9 p^2+18$
Using the Identity $x^2+(a+b) x+a b=(x+a)(x+b)$
We find $a b=18$ and $a+b=9$
From this, we must obtain $a$ and $b$.
If $a b=18$, it means that $a$ and $b$ are factor of 18 .
Let us try $a=6$ and $b=3$,
For these values $a+b=6+3=9$ exactly as required.
$
\begin{aligned}
\therefore \quad p^4+9 p^2+18 & =p^4+6 p^2+3 p^2+18 \\
& =p^2\left(p^2+6\right)+3\left(p^2+6\right) \\
& =\left(p^2+3\right)\left(p^2+6\right)
\end{aligned}
$
Hence, Salma shows the correct steps of factorisation.

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Question 25 Marks

Perform the following divisions.
(i) $\left(3 p q r-6 p^2 q^2 r^2\right)÷ 3 p q$
(ii) $\left(a x^3-b x^2+c x\right)÷ (-d x)$
(iii) $\left(x^3 y^3+x^2 y^3-x y^4 + x y\right)÷x y$
(iv) $(-q r x y+p r y z-n y z)÷ (-x y z)$

Answer

(i) We have, $\left(3 p q r-6 p^2 q^2 r^2\right)÷ 3 p q$
$
\begin{array}{l}
=\frac{3 p q r-6 p^2 q^2 r^2}{3 p q}=\frac{3 p q r}{3 p q}-\frac{6 p^2 q^2 r^2}{3 p q} \\
=r-\frac{2 \times 3 \times p \times p \times q \times q \times r \times r}{3 \times p \times q}=r-2 p q r^2
\end{array}
$
(ii) We have, $\left(a x^3-b x^2+c x\right) \div(-d x)=\frac{a x^3-b x^2+c x}{-d x}$
$
\begin{array}{l}
=\frac{a x^3}{-d x}+\frac{b x^2}{d x}+\frac{c x}{-d x} \\
=\frac{a \times x \times x \times x}{-d \times x}+\frac{b \times x \times x}{d \times x}+\frac{c \times x}{-d \times x}=-\frac{a}{d} x^2+\frac{b}{d} x-\frac{c}{d}
\end{array}
$
(iii) We have, $\left(x^3 y^3+x^2 y^3-x y^4+x y\right) \div x y$
$
\begin{array}{l}
=\frac{x^3 y^3+x^2 y^3-x y^4+x y}{x y} \\
=\frac{x^3 y^3}{x y}+\frac{x^2 y^3}{x y}-\frac{x y^4}{x y}+\frac{x y}{x y} \\
=\frac{x \times x \times x \times y \times y \times y}{x \times y}+\frac{x \times x \times y \times y \times y}{x \times y}
\end{array}
$
$-\frac{x \times y \times y \times y \times y}{x \times y}+\frac{x \times y}{x \times y}$
$
=x^2 y^2+x y^2-y^3+1
$
(iv) We have,
$\begin{array}{l}
(-q r x y+p r y z-r x y z)÷(-x y z) \\
=\frac{-q r x y+p r y z-r x y z}{-x y z} \\
=\frac{-q r x y}{-x y z}+\left(\frac{p r y z}{-x y z}\right)-\left(\frac{r x y z}{-x y z}\right)=\frac{q r}{z}-\frac{p r}{x}+r
\end{array}
$

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Question 35 Marks

Carry out the following divisions.
(i) $51 x^3 y^2 z÷17 x y z$
(ii) $76 x^3 y z^3÷19 x^2 y^2$
(iii) $17 a b^2 c^3÷\left(-a b c^2\right)$
(iv) $-121 p^3 q^3 r^3÷\left(-11 x y^2 z^3\right)$

Answer

(i) We have,
$
\begin{aligned}
51 x^3 y^2 z÷17 x y z & =\frac{51 x^3 y^2 z}{17 x y z} \\
& =\frac{17 \times 3 \times x \times x \times x \times y \times y \times z}{17 \times x \times y \times z}=3 x^2 y
\end{aligned}
$
(ii) We have,
$
\begin{array}{l}
76 x^3 y z^3÷19 x^2 y^2=\frac{76 x^3 y z^3}{19 x^2 y^2} \\
=\frac{4 \times 19 \times x \times x \times x \times y \times z \times z \times z}{19 \times x \times x \times y \times y}=\frac{4 z^3}{y}
\end{array}
$
(iii) We have,
$
\begin{aligned}
17 a b^2 c^3÷\left(-a b c^2\right) & =\frac{17 a b^2 c^3}{-a b c^2} \\
& =\frac{17 \times a \times b \times b \times c \times c \times c}{-a \times b \times c \times c}=-17 b c
\end{aligned}
$
(iv) We have,
$
\begin{array}{l}
-121 p^3 q^3 r^3÷\left(-11 x y^2 z^3\right)=\frac{-121 p^3 q^3 r^3}{-11 x y^2 z^3} \\
=\frac{-11 \times 11 \times p \times p \times p \times q \times q \times q \times r \times r \times r}{-11 \times x \times y \times y \times z \times z \times z}=\frac{11 p^3 q^3 r^3}{x y^2 z^3}
\end{array}
$

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Question 45 Marks

Factorise the following, using the identity $\left(a^2-b^2\right)=(a-b)(a+b)$.
(i) $4 x^2-25 y^2$
(ii) $\frac{2 p^2}{25}-32 q^2$
(iii) $\frac{x^3 y}{9}-\frac{x y^3}{16}$
(iv) $16 x^4-81$

Answer

(i) We have, $4 x^2-25 y^2=(2 x)^2-(5 y)^2$
$
=(2 x-5 y)(2 x+5 y)
$
(ii) We have,
$
\begin{aligned}
\frac{2 p^2}{25}-32 q^2 & =2\left(\frac{p^2}{25}-16 q^2\right)=2\left[\left(\frac{p}{5}\right)^2-(4 q)^2\right] \\
& =2\left(\frac{p}{5}-4 q\right)\left(\frac{p}{5}+4 q\right)
\end{aligned}
$
(iii) We have,
$
\begin{aligned}
\frac{x^3 y}{9}-\frac{x y^3}{16} & =x y\left(\frac{x^2}{9}-\frac{y^2}{16}\right)=x y\left[\left(\frac{x}{3}\right)^2-\left(\frac{y}{4}\right)^2\right] \\
& =x y\left(\frac{x}{3}-\frac{y}{4}\right)\left(\frac{x}{3}+\frac{y}{4}\right)
\end{aligned}
$
(iv) We have,
$
\begin{aligned}
16 x^4-81 & =\left(4 x^2\right)^2-(9)^2 \\
& =\left(4 x^2-9\right)\left(4 x^2+9\right) \\
& =\left[(2 x)^2-(3)^2\right]\left(4 x^2+9\right) \\
& =(2 x-3)(2 x+3)\left(4 x^2+9\right)
\end{aligned}
$

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Question 55 Marks

Factorise the following, using the identity $\left(a^2-2 a b+b^2\right)=(a-b)^2$.
(i) $y^2-14 y+49$
(ii) $\frac{x^2}{4}-2 x+4$
(iii) $a^2 y^3-2 a b y^2+b^2 y$
(iv) $9 y^2-4 x y+\frac{4 x^2}{9}$

Answer

(i) $\begin{array}{l} y^2-14 y+49=y^2-2 \times 7 \times y+7^2 \\ =(y-7)^2=(y-7)(y-7) \\ \text { (ii) } \begin{aligned} \frac{x^2}{4}-2 x+4 & =\left(\frac{x}{2}\right)^2-2 \times \frac{x}{2} \times 2+2^2 \\ & =\left(\frac{x}{2}-2\right)^2=\left(\frac{x}{2}-2\right)\left(\frac{x}{2}-2\right)\end{aligned}\end{array}$
(iii)
$
\begin{aligned}
a^2 y^3-2 a b y^2+b^2 y & =y\left(a^2 y^2-2 a b y+b^2\right) \\
& =y\left[(a y)^2-2 \times(a y) \times b+b^2\right] \\
& =y(a y-b)^2=y(a y-b)(a y-b)
\end{aligned}
$
(iv)
$
\begin{aligned}
9 y^2-4 x y+\frac{4 x^2}{9} & =(3 y)^2-2 \times 3 y \times \frac{2 x}{3}+\left(\frac{2 x}{3}\right)^2 \\
& =\left(3 y-\frac{2 x}{3}\right)^2 \\
& =\left(3 y-\frac{2 x}{3}\right)\left(3 y-\frac{2 x}{3}\right)
\end{aligned}
$

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Question 65 Marks

Factorise the following expressions.
(i) $a x^3-b x^2+c x$
(ii) $x^3 y^2+x^2 y^3-x y^4+x y$
(iii) $a^3+a^2+a+1$
(iv) $a x^2 y-b x y z-a x^2 z+b x y^2$

Answer

(i) $a x^3-b x^2+c x=x\left(a x^2-b x+c\right)$
(ii) $x^3 y^2+x^2 y^3-x y^4+x y=x y\left(x^2 y+x y^2-y^3+1\right)$
(iii) $a^3+a^2+a+1=a^2(a+1)+1(a+1)=(a+1)\left(a^2+1\right)$
(iv) $a x^2 y-b x y z-a x^2 z+b x y^2$
$\begin{array}{l}=x\left(a x y-b y z-a x z+b y^2\right) \\ =x\left(a x y-a x z+b y^2-b y z\right)\end{array}\qquad \text {[regrouping the terms]}$
$\begin{array}{l}=x[a x(y-z)+b y(y-z)] \\ =x(y-z)(a x+b y)\end{array}$

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Question 75 Marks

Write the greatest common factor in each of the following terms.
(i) $-18 a^2, 108 a$
(ii) $3 x^2 y, 18 x y^2,-6 x y$
(iii) $l^2 m^2 n, l m^2 n^2, l^2 m n^2$
(iv) $63 p^2 a^2 r^2 s,-9 p q^2 r^2 s^2, 15 p^2 q r^2 s^2,-60 p^2 a^2 r s^2$

Answer

(i)
$
\begin{array}{l}
-18 a^2=-18 \times a \times a \text { and } 108 a=18 \times 6 \times a \\
\therefore G C F=18 a .
\end{array}
$
(ii)
$
\begin{aligned}
3 x^2 y & =3 \times x \times x \times y \\
18 x y^2 & =3 \times 6 \times x \times y \times y \text { and }-6 x y=3 \times(-2) \times x \times y \\
\therefore GCF & =3 x y
\end{aligned}
$
(iii)
$
\begin{array}{l}
l^2 m^2 n=l \times l \times m \times m \times n \\
l m^2 n^2=l \times m \times m \times n \times n \text { and } \\
l^2 m n^2=l \times l \times m \times n \times n \\
\therefore GCF=l m n
\end{array}
$
(iv)
$
\begin{array}{l}
63 p^2 a^2 r^2 s=3 \times 3 \times 7 \times p \times p \times a \times a \times r \times r \times s \\
-9 p q^2 r^2 s^2=-3 \times 3 \times p \times q \times q \times r \times r \times s \times s \\
15 p^2 q r^2 s^2=3 \times 5 \times p \times p \times q \times r \times r \times s \times s \\
-60 p^2 a^2 r s^2=-2 \times 2 \times 3 \times 5 \times p \times p \times a \times a \times r \times s \times s \\
\therefore \text { GCF }=3 p r s
\end{array}
$

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MATHS 5 Marks Questions

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