5 Marks Questions - Maths STD 8 Questions - Vidyadip

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29 questions · self-marked practice — reveal the answer and mark yourself.

Question 15 Marks

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

Answer

The given expression is Assuming $x = 2a -b$, we have:
$(2a - b)^2+ (2a - b) - 8 = x^2+ 2x - 8$
$($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., $1 × (-8) = -8$
Now,
$(-2) + 4 = 2$ And $(-2) × 4 = -8$
Replacing the middle term $2x by -2x + 4x$, we have:
$(2a - b)^2+ (2a - b) - 8 = x^2- 2x + 4x - 8$
$= (x^2 - 2x) + (4x - 8)$
$= x(x - 2) + 4(x - 2)$
$= (x - 2)(x + 4)$
Replacing $x$ by $2a - b$, we get:
$(x + 4)(x - 2) = (2a - b + 4)(2a - b - 2)$

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Question 25 Marks

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

Answer

The given expression is $(x - 2y)^2- 5(x - 2y) + 6$
Assuming a $= x - 2y,$ we have:
$(x - 2y)^2- 5(x - 2y) + 6 = a^2- 5a + 6$
$($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., $1 × 6= 6$
Clesrly,
$(-2) + (-3) = (-5)$ And $(-2) \times (-3) = 6$
Replacing the middle term -$5a$ by $-2a -3a$,
we have: $(x - 2y)^2- 5(x - 2y) + 6 = a^2- 2a - 3a + 6$
$= (a^2- 2a) - (3a - 6)$
$= a (a - 2) - 3(3a - 6)$
$= (a - 2)(a - 3)$
Replacing a by $(x - 2y)$,
we get: $(a - 3)(a - 2) = (x - 2y - 3)(x - 2y - 2)$

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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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Question 45 Marks

Resolve of the folloeing quadratic equation trinomials into factor:
$15 x^2-16 x y z-15 y^2 z^2$

Answer

The given expression is $15 x^2-16 x y z-15 y^2 z^2$
(co-effcient of $x^2=15$, co-effcient of $x=16 y z$ and the constant term $=-15 y^2$ )
We will split the co-efficient of $x$ into two parts such that thier sum in is $-16 y z$ and thier product equals to the product of the co-efficient of $x^2$ and the constant term, i.e., $15 \times\left(-15 y^2 z^2\right)=-225 y^2 z^2$
Now,
$(-25 y z)+(9 y z)=(-16 y z) \text { And }(-25 y z) \times(9 y z)=-225 y^2 z^2$
Replacing the middle term $-16 x y z$ by $-25 x y z+9 x y z$, we have:
$ 15 x^2-16 x y z-15 y^2 z^2=15 x^2-25 x y z+9 x y z-15 y^2 z^2 $
$ =\left(15 x^2-25 x y z\right)+\left(9 x y z-15 y^2 z^2\right) $
$ =5 x(3 x-5 y z)+3 y z(3 x-5 y z) $
$ =(3 x-15 y z)(5 x+3 y z)$

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Question 55 Marks

Resolve each of the following quadratic trinomial into factor:
$2x^2+ 5x + 3$

Answer

The given expression is $2x^2+ 5x + 3$
$($Co-efficient of $x^2= 2,$ co-efficient of $x = 5$ and the constant term $= 3)$
We will split the co-efficient of x into two number parts such that their sum is $5$ and their product equals to the product of the co-efficient of $x^2$ and constant term, i.e.,$ 2 × 3 = 6$
Now,
$2 + 5 = 5$ And $2 × 3 = 6$
Replacing the middle term $5x$ by $2x + 3x$, we have:
$ 2 x^2+5 x+3=2 x^2+2 x+3 x+3 $
$=\left(2 x^2+2 x\right)+(3 x+3) $
$=2 x(x+1)+3(x+1) $
$ =(2 x+3)(x+1)$

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Question 65 Marks

Solve:
$p^2+ 6p + 8$

Answer

$p^2+ 6p + 8$
$=\text{p}^2+6\text{p}+\Big(\frac{6}{2}\Big)^2-\Big(\frac{6}{2}\Big)^2 [$Adding and subtracting $\Big(\frac{6}{2}\Big)^2,$ that is $3^2]$
$ =p^2+6 p+3^2-3^2+8 $
$ =p^2+2 \times p \times 3+3^2-9+8 $
$ =p^2+2 \times p \times 3+3^2-1 $
$ =(p+3) 2-12[\text { completing the square }] $
$ =[(p+3)-1][(p+3)+1] $
$ =(p+3-1)(p+3+1) $
$ =(p+2)(p+4) $

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Question 75 Marks

Resolve each of the following quadratic trinomial into factor: $2x^2- 3x - 2$

Answer

The given expression is $2x^2- 3x - 2$
(Co-efficient of $x^2= 2,$ co-efficient of $x = -3$ and the constant term $= -2)$
We will split the co-efficient of x into two number parts such that their sum is $-3$ and their product equals to the product of the co-efficient of $x^2$ and constant term, i.e., $2 \times (-2) = -4$
Now,
$(-4) + 1 = -3$ And $(-4) \times 1 = -4$
Replacing the middle term $3x$ by $-4x + x$, we have:
$ 2 x^2-3 x-2=2 x^2-4 x+x-2 $
$ =\left(2 x^2-4\right)+(x-2)$
$ =2 x(x-2)+1(x-2) $
$ =(x-2)(2 x+1) $

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Question 85 Marks

Resolve each of the following quadratic trinomial into factor:
$7 x-6 x^2+20$

Answer

The given expression is $7 x-6 x^2+20$
(Co-efficient of $x^2=6$, co-efficient of $x=7$ and the constant term $=20$ )
We will split the co-efficient of $x$ into two number parts such that their sum is $-19$ and their product equals to the product of the co-efficient of $x^2$ and constant term, i.e., $(-6) \times 20=-120$
Now,
$(15)+(-8)=7 \text { And }(15) \times(-8)=120$
Replacing the middle term $7 x$ by $15 x-21 x$, we have:
$ 7 x-6 x^2+20=-6 x^2+15 x-8 x+20 $
$ =\left(-6 x^2+15 x\right)+(-8 x+20) $
$ =3 x(-2 x+5)+4(-2 x+5)$
$ =(-2 x+5)(3 x+4)$

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Question 95 Marks

Resolve each of the following quadratic trinomial into factor: $3 x^2+10 x+3$

Answer

The given expression is $3 x^2+10 x+3$
(Co-efficient of $x^2=3$, co-efficient of $x=10$ and the constant term $=3$ )
We will split the co-efficient of $x$ into two number parts such that their sum is $-3$ and their product equals to the product of the co-efficient of $x^2$ and constant term, i.e., $3 \times 3=9$
Now,
$9+1=9 \text { And } 9 \times 1=9$
Replacing the middle term $10 x$ by $9 x+x$, we have:
$ 3 x^2+10 x+3=3 x^2+9 x+x+3$
$ =\left(3 x^2+9 x\right)+(x+3)$
$ =3 x(x+3)+1(x+3)$
$ =(x+3)(3 x+1)$

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Question 105 Marks

Resolve each of the following quadratic trinomial into factor:
$ 3 x^2+22 x+35$

Answer

The given expression is $3x^2+ 22x + 35$
$($Co-efficient of $x^2= 3,$ co-efficient of $x = 22$ and the constant term $= 35)$
We will split the co-efficient of $x$ into two number parts such that their sum is $-19$ and their product equals to the product of the co-efficient of $x^2$ and constant term, i.e., $ 3 × 35 = 105$
Now,
$(15) + (7) = 22$ And $(15) × (7) = 105$
Replacing the middle term $22x$ by $15x\ 7x,$ we have:
$ 3 x^2+22 x+35$
$=3 x^2+15 x+7 x+35 $
$ =\left(3 x^2+15 x\right)+(7 x+35) $
$ =3 x(x+5)+7(x+5) $
$ =(x+5)(3 x+7)$

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Question 115 Marks

Solve:
$x^2+ 12x + 20$

Answer

$x^2+ 12x + 20$
$=\text{x}^2 +12\text{x}+\Big(\frac{12}{2}\Big)^2-\Big(\frac{12}{2}\Big)^2+20$ $\Big[$Adding and subtractiong $\Big(\frac{12}{2}\Big)^2,$ that is $6^2\Big]$
$ =x^2+12 x+6^2-6^2+20 $
$ =(x+6)^2-16[\text { comleting the square }] $
$ =(x+6)^2-4^2 $
$ =[(x+6)-4][(x+6)+4] $
$ =(x+6-4)(x+6+4) $
$ =(x+2)(x+10)$

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Question 125 Marks

Resolve each of the following quadratic trinomial into factor:
$7x^2- 19x - 6$

Answer

The given expression is $7x^2- 19x - 6$
$($Co-efficient of $x^2= 7$, co-efficient of $x = -19$ and the constant term $= -6)$
We will split the co-efficient of $x$ into two number parts such that their sum is $-3$ and their product equals to the product of the co-efficient of $x^2$ and constant term, i.e., $7 × (-6) = 9$
Now,
$(-21) + 2 = 19$ And $(-21) × 2 = -42$
Replacing the middle term $-19x$ by $-21x + x,$ we have:
$ 7 x^2-19 x-6=7 x^2-21 x+2 x-6 $
$ =\left(7 x^2-21 x\right)+(2 x-6) $
$ =7 x(x-3)+2(x-3)$
$ =(x-3)(7 x-2)$

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Question 135 Marks

Resolve each of the following quadratic trinomial into factor:
$3+23 y-8 y^2$

Answer

The given expression is $3+23 y-8 y^2$
$($Co-efficient of $y^2=8$, co-efficient of $y = 23$ and the constant term $= 3)$
We will split the co-efficient of $y$ into two number parts such that their sum is $-31$ and their product equals to the product of the co-efficient of $y^2$ and constant term, i.e., $(-8) × 3 = -24$
Now,
$(-1) + 24 = 23$ And $(-1) \times 24 = -24$
Replacing the middle term $23y$ by $-y + 24y$,
we have:
$ 3+23 y-8 y^2=-8 y^2-y+24 y+3 $
$ =\left(-8 y^2-y\right)+(24 y+3) $
$ =-y(8 y+1)+3(8 y+1) $
$ =(8 y+1)(y+z)$

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Question 145 Marks

Resolve of the folloeing quadratic equation trinomials into factor: $6a^2+ 17ab - 3b^2$

Answer

The given expression is $6a^2+ 17ab - 3b^2$
(co-effcient of $a^2= 6$, co-effcient of $a = 17b$ and the constant term $= -3b^2$)
We will split the co-efficient of $x$ into two parts such that thier sum in is $17b$ and thier product equals to the product of the co-efficient of $a^2$ and the constant term, i.e., $6 \times (-3b^2) = 18b^2$
Now,
$(18b) + (-b) = 17b$ And $(18b) \times (-b) = - 18b^2$
Replacing the middle term $17ab$ by $-ab + 18ab$, we have:
$6 a^2+17 a b-3 b^2=6 a^2-a b+18 a b-3 b^2$
$=\left(6 a^2-a b\right)+\left(18 a b-3 b^2\right)$
$=a(6 a-b)+3 b(6 a-b)$
$=(a+3 b)(6 a-b)$

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Question 155 Marks

Resolve each of the following quadratic trinomial into factor:
$7x - 6 - 2x^2$

Answer

The given expression is $7x - 6 - 2x^2$
(Co-efficient of $x^2= -2,$ co-efficient of $x = 7$ and the constant term $= -6$)
We will split the co-efficient of $x$ into two number parts such that their sum is $7$ and their product equals to the product of the co-efficient of $x^2$ and constant term, i.e., $(-2) × (-6) = 12$
Now,
$4 + 3 = 7$ And $4 \times 3 = 12$
Replacing the middle term $7x$ by $4x + 3x,$ we have:
$ 7 x-6-2 x^2 $
$ =\left(-2 x^2+4 x\right)+(3 x-6) $
$ =2 x(2-x)-3(2-x) $
$ =(2 x-3)(2-x) $

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Question 165 Marks

Resolve each of the following quadratic trinomial into factor:
$11x^2- 54x + 63$

Answer

The given expression is $11x^2- 54x + 63$
(Co-efficient of $x^2= 11$, co-efficient of $x = 54$ and the constant term $= 63$)
We will split the co-efficient of x into two number parts such that their sum is $-19$ and their product equals to the product of the co-efficient of $x^2$ and constant term, i.e., $11 \times 36 = 693$
Now,
$(-33)+(-21)=-54 \text { And }(-33) \times(-21)=693$
Replacing the middle term $-54 x$ by $-33 x+-21 x$, we have:
$ 11 x^2-54 x+63=11 x^2-33 x-21 x+63$
$ =\left(11 x^2-33 x\right)+(-21 x+63)$
$ =11 x(x-3)-21(x-3) $
$=(x-3)(11 x-21)$

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Question 175 Marks

Resolve each of the following quadratic trinomial into factor:
$28 - 31x - 5x^2$

Answer

The given expression is $28 - 31x - 5x^2$
(Co-efficient of $x^2= 5,$ co-efficient of $x = -31$and the constant term $= 28$)
We will split the co-efficient of x into two number parts such that their sum is$ -31$ and their product equals to the product of the co-efficient of $x^2$ and constant term, i.e., $(-5) \times (28) = -140$
Now,
$(-35)+4=-31 \text { And }(-35) \times 4=-140$
Replacing the middle term $-31 x$ by $-35 x+4 x$,
we have: $28-31 x-5 x^2=-5 x^2-35 x+4 x+28$
$=\left(-5 x^2-35 x\right)+(4 x+28) $
$ =-5 x(x+7)+4(x+7)$
$=(4-5 x)(x+7)$

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Question 185 Marks

Solve:
$a^2+ 2a - 3$

Answer

$a^2+ 2a - 3$
$=\text{a}^2+2\text{a}+\Big(\frac{2}{2}\Big)^2-\Big(\frac{2}{2}\Big)^2-3$ $\Big[$Adding and subtracting $\Big(\frac{2}{2}\Big)^2,$ that is $ 1^2\Big]$
$= a2 + 2a + 1 - 1 - 3$
$= (a + 1)2 - 4 [$Comleting the square$]$
$= (a + 1)2 - 22$
$= [(a + 1) - 2][(a + 1) + 2]$
$= (a + 1 - 2)(a + 1 + 2)$
$= (a - 1)(a + 3)$

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Question 195 Marks

Resolve each of the following quadratic trinomial into factor:
$12x^2- 17xy + 6y^2$

Answer

The given expression is $12x^2- 17xy + 6y^2$
(Co-efficient of $x^2= 12,$ co-efficient of $x = 17y$ and the constant term $= 6y^2$)
We will split the co-efficient of $x$ into two number parts such that their sum is $-17y$ and their product equals to the product of the co-efficient of $x^2$ and constant term, i.e., $12 × 6y^2= 72y^2$
Now,
$(-9 y)+(-8 y)=17 y \text { And }(-9 y) \times(-8 y)=72 y^2$
Replacing the middle term $-17 x y$ by $-9 x y-8 x y$,
we have:
$ 12 x^2-17 x y+6 y^2=12 x^2-9 x y-8 x y+6 y^2$
$ =\left(12 x^2-9 x y\right)-\left(8 x y+6 y^2\right)$
$=3 x(4 x-3 y)-2 y(4 y-3 y) $
$ =(4 x-3 y)(3 x-2 y)$

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Question 205 Marks

Solve:
$4x^2- 12x + 5$

Answer

$4x^2- 12x + 5$
$=\Big(\text{x}^2 - 3\text{x} +\frac{5}{4}\Big)$ [Making the co-efficient of $x^2= 1$]
$=4\Big[\text{x}^2-3\text{x}+\Big(\frac{3}{2}\Big)^2-\Big(\frac{3}{2}\Big)^2+ \frac{5}{4}\Big]$ $\Big[$Adding and subtracting $\Big(\frac{3}{2}\Big)^2\Big]$
$=4\Big[\Big(\text{x}-\frac{3}{2}\Big)^2-\frac{9}{4}+\frac{5}{4}\Big]$ [Compliting the suare]
$=4\Big[\Big(\text{x}-\frac{3}{2}\Big)^2 -1\Big]$
$=4\Big[\Big(\text{x}-\frac{3}{2}\Big)-1\Big]\Big[\Big(\text{x}-\frac{3}{2}\Big)+1\Big]$
$=4\Big(\text{x}-\frac{3}{2}-1\Big)\Big(\text{x}-\frac{3}{2}+1\Big)$
$=4\Big(\text{x}-\frac{5}{2}\Big)\Big(\text{x}-\frac{1}{2}\Big)$
$=(2\text{x}-5)(2\text{x}-1)$

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Question 215 Marks

Factories:
$(a + 7)(a - 10) + 16$

Answer

$(a + 7)(a - 10) + 16$
$ =a^2-10 a+7 a-70+16$
$=a^2-3 a-54$
To factories $a^2-3 a-54$, we will find two numbers $p$ and $q$ such that $p + q = -3$ and $pq = -54$
Now,
$6 + (-9) = -3$ And $6 x (-9) = -54$
Splitting the middle term $-3a$ in the given quadratic as $-9 + 6a$,
we get:
$a^2-3 a-54=a^2-9 a+6 a-54$
$ =\left(a^2-9 a\right)+(6 a-5) $
$ =a(a-9)+6(a-9) $
$ =(a+6)(a-9)$

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Question 225 Marks

Resolve each of the following quadratic trinomial into factor:
$6x^2- 5xy + 6y^2$

Answer

The given expression is $6x^2- 5xy + 6y^2$
(Co-efficient of $x^2= 6$, co-efficient of $x = 5y$ and the constant term $= 6y^2$)
We will split the co-efficient of $x$ into two number parts such that their sum is $-17y$ and their product equals to the product of the co-efficient of $x^2$ and constant term, i.e., $6 \times (-6y^2) = 36y^2$
Now,
$(-9 y)+(4 y)=-5 y \text { And }(-9 y) \times(4 y)=-36 y^2$
$\text { Replacing the middle term }-5 x y \text { by }-9 x y+4 x y \text {, we have: }$
$ 6 x^2-5 x y+6 y^2=6 x^2-9 x y+4 x y-6 y^2 $
$ =\left(6 x^2-9 x y\right)+\left(4 x y-6 y^2\right) $
$ =3 x(2 x-3 y)+2 y(2 x-3 y) $
$ =(2 x-3 y)(3 x+2 y)$

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Question 235 Marks

Resolve of the folloeing quadratic equation trinomials into factor:
$36a^2+ 12abc - 15b^2c^2$

Answer

The given expression is $36a^2+ 12abc - 15b^2c^2$
(co-effcient of $a^2= 36,$ co-effcient of $a = 12bc$ and the constant term $= -15b^2$)
We will split the co-efficient of $x$ into two parts such that thier sum in is $17\ b$ and thier product equals to the product of the co-efficient of $a^2$ and the constant term, i.e., $36 × (-15b^2c^2) = -540b^2c^2$
Now, $(-18bc) + (30bc) = 12bc$ And $(18bc) × (30bc) = -540b^2c^2$
Replacing the middle term $17ab- by -ab + 18ab$,
we have:$36a^2+ 12abc - 15b^2c^2= 36a^2- 18abc + 30abc - 15b^2c^2$
$= (36a^2- 18abc) + (30abc - 15b^2c^2)$
$= 18a(2a - bc) + 15bc(2a - bc)$
$= 3(6a + 5abc)(2a - bc)$

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Question 245 Marks

Factories: $40 + 3x - x^2$

Answer

We have:
$40 + 3x - x^2$
$= -(x^2- 3x - 40)$
To factories $(x^2- 3x - 40),$ we fill find two number p and q such $p + q = -3$ and $pq = -40$
Now,
$5 + (-8) = -3$ and $5 x (-8) = -40$
$= -(x^2+ 5x - 8x - 40)$
$= -[x(x + 5) - 8(x + 40)]$
$= -(x - 8)(x + 5)$
$= (x + 5)(-x + 8)$

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Question 255 Marks

Solve:
$4y^2+ 12y + 5$

Answer

$4y^2+ 12y + 5$
$4\Big(\text{y}^2=3\text{y}+\frac{5}{4}\Big)$ [ making the co-efficient of $y^2$]
$4\Big[\text{y}^2+3\text{y}+\Big(\frac{3}{2}\Big)^2-\Big(\frac{3}{2}\Big)^2=\frac{5}{4}$ $\Bigg[$Adding and subtracting $\Big(\frac{3}{2}\Big)^2\Bigg]$
$=4\Big[(\text{y}+\frac{3}{2})^2-\frac{9}{4}+\frac{5}{4}\Big]$
$=4\Big[\Big(\text{y}+\frac{3}{2}\Big)^2-1\Big]$ [ completing the square]
$=4\Big[\Big(\text{y}+\frac{3}{2}-1\Big)\Big]\Big[\Big(\text{y}+\frac{3}{2}+1\Big)\Big]$
$=4\Big(\text{y}+\frac{3}{2}-1\Big)\Big(\text{y}+\frac{3}{2}-1\Big)$
$=4\Big(\text{y}+\frac{3}{2}\Big)\Big(\text{y}+\frac{3}{2}\Big)$
$=(2\text{y}+1)(2\text{y}+5)$

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Question 265 Marks

Resolve each of the following quadratic trinomial into factor:
$14x^2+ 11xy - 15y^2$

Answer

The given expression is $14x^2+ 11xy - 15y^2$
(Co-efficient of $x^2=14$, co-efficient of $x=11 y$ and the constant term $=-15 y^2$ )
We will split the co-efficient of $x$ into two number parts such that their sum is $11 y$ and their product equals to the product of the co-efficient of $x^2$ and constant term, i.e., $14 \times\left(-15 y^2\right)=-210 y^2$
Now,
$(21 y)+(-10 y)=11 y \text { And }(21 y) \times(-10 y)=-210 y^2$
Replacing the middle term $-11 x y$ by $-10 x y+21 x y$, we have:
$ 14 x^2+11 x y-15 y^2=14 x^2-10 x y+21 y-15 y^2 $
$ =\left(14 x^2-10 x y\right)+\left(21 x y-15 y^2\right) $
$ =2 x(7 x-5 y)+3 y(7 x-5 y) $
$ =(7 x-5 y)(2 x+3 y)$

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Question 275 Marks

Resolve each of the following quadratic trinomial into factor:
$6x^2- 13xy + 2y^2$

Answer

The given expression is $6x^2- 13xy + 2y^2$
(Co-efficient of $x^2= 6,$ co-efficient of $x = -13y$ and the constant term $= 2y^2$) We will split the co-efficient of $x$ into two number parts such that their sum is $-13y$ and their product equals to the product of the co-efficient of $x^2$ and constant term, i.e., $6 \times (2y^2) = 12y^2$
Now,
$(-12 y)+(-y)=-13 y \text { And }(-12 y) \times(-y)=12 y^2$
Replacing the middle term $-13 x y$ by $-12 x y-x y$,
we have: $6 x^2-13 x y+2 y^2=6 x^2-12 x y-x y+2 y^2 $
$ =\left(6 x^2-12 x y\right)-(x y-2 y)$
$ =6 x(x-2 y)-y(x-2 y) $
$ =(x-2 y)(6 x-y)$

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Question 285 Marks

Solve:
$(z - 6)(z + 2)$

Answer

$= z^2- 4z - 12$
$\Big[$Adding and subtracting $\Big(\frac{4}{2}\Big)^2\Big]$
$=\text{z}^2-4\text{z}+\Big(\frac{4}{2}\Big)^2-\Big(\frac{4}{2}\Big)^2-12$
$=\text{z}^2-4\text{z}=(2)^2-(2)^2-12$
$=(\text{z}-2)^2-16$
Comleting the squares
$=(\text{z}-2)^2-(4)^2$
$=[(\text{z}-2)-4][(\text{z}-2)+4]$
$=(\text{z}-6)(\text{z}+2)$

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Question 295 Marks

Solve:
$a^2- 14a - 51$

Answer

$a^2- 14a - 51$
$=\text{a}^2-14\text{a}+\Big(\frac{14}{2}\Big)^2-\Big(\frac{14}{2}\Big)^2-51$ $\Big[$Adding and subtracting $\Big(\frac{14}{2}\Big)^2,$ that is $7^2\Big]$
$ =a^2-14 a+7^2-7^2-51 $
$ =(a-7)^2-100[\text { comleting the square }] $
$ =(a-7)^2-10^2 $
$ =[(a-7)-10][(a-7)+10] $
$ =(a-7-10)(a-7+10) $
$ =(a-17)(a+3)$

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