5 Mark Question

Question 15 Marks

Resolve of the folloeing quadratic equation trinomials into factor:
$(2 a-b)^2+(2 a-b)-8$

Answer

The given expression is Assuming $x =2 a - b$, we have: $(2 a - b )^2+(2 a - b )-8= x ^2+2 x -8$
(co-effcient of $x^2=1$, co-effcient of $x=2$ and the constant term $=-8$ )
We will split the co-efficient of $x$ into two parts such that thier sum in is 2 and thier product equals to the product of the co-efficient of $x^2$ and the constant term, i.e., $1 \times(-8)=-8$
Now,
$(-2)+4=2 \text { And }(-2) \times 4=-8$
Replacing the middle term $2 x$ by $-2 x+4 x$, we have:
$(2 a-b)^2+(2 a-b)-8=x^2-2 x+4 x-8$
$=\left(x^2-2 x\right)+(4 x-8)$
$=x(x-2)+4(x-2)$
$=(x-2)(x+4)$
Replacing $x$ by $2 a - b$, we get:
$(x+4)(x-2)=(2 a-b+4)(2 a-b-2)$

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Question 25 Marks

Resolve of the folloeing quadratic equation trinomials into factor: $(x-2 y)^2-5(x-2 y)+6$

Answer

The given expression is $(x-2 y)^2-5(x-2 y)+6$ Assuming $a=x-2 y$, we have:
$(x-2 y)^2-5(x-2 y)+6=a^2-5 a+6$
(co-effcient of $a^2=1$, co-effcient of $a=-5$ and the constant term $=6$ ) We will split the co-efficient of $x$ into two parts such that thier sum in is -5 and thier product equals to the product of the co-efficient of $a^2$ and the constant term, i.e., $1 \times 6=6$ Clesrly, $(-2)+(-3)=(-5)$ And $(-2) \times(-3)=6$ Replacing the middle term -5 a by $-2 a -3 a$, we have: $( x$
$-2 y)^2-5(x-2 y)+6=a^2-2 a-3 a+6$
$=\left(a^2-2 a\right)-(3 a-6)$
$=a(a-2)-3(3 a-6)$
$=(a-2)(a-3)$
Replacing a by ( $x-2 y$ ), we get:
$(a-3)(a-2)=(x-2 y-3)(x-2 y-2)$

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Question 35 Marks

Factories: $\left(a^2-54\right)^2-36$

Answer

$\left(a^2-54\right)^2-36=\left(a^2-5 a\right)^2-6^2$
$=\left[\left(a^2-5 a\right)-6\right]\left[\left(a^2-5 a\right)+6\right]$
$=\left(a^2-5 a-6\right)\left(a^2-5 a+6\right)$
In order to factories $a^2-5 a-6$, we will find two number $p$ and $q$ such that $p+q=-5$ and $p q=-6$
Now,
$(-6)+1=-5 \text { And }(-6) \times 1=-6$
Splitting the middle term -5 in the given quadratic as $-6+a$, we get:
$a^2-5 a-6=a^2-6 a+a-6$
$=\left(a^2-6 a\right)+(a-6)$
$=a(a-6)+(a-6)$
$=(a+1)(a-6)$
Now, In order to factories $a^2-5 a+6$, we will find two numbers $p$ and $q$ such that $p+q=-5$ and $p q=6$ Clearly,
$(-2)+(-3)=-5 \text { and }(-2) \times(-3)=6$
Splitting the middle term -5 in the quadratic as $-2 a-3 a$, we get:
$a^2-5 a+6=a^2-2 a-3 a+6$
$=\left(a^2-2 a\right)-(3 a-6)$
$=a(a-2)-3(a-2)$
$=(a-3)(a-2)$
$\therefore\left(a^2-5 a-6\right)\left(a^2-5 a+6\right)$
$=(a-6)(a+1)(a-3)(a-2)$
$=(a+1)(a-2)(a-3)(a-6)$

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Maths 5 Mark Question

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