4 Mark Question - Maths STD 8 Questions - Vidyadip

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4 Mark Question

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

Question 14 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, $(-25 y z)+(9 y z)=(-16 y z)$ 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 24 Marks

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

Answer

The given expression is $2 x^2+5 x+3$
(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., $2 \times 3=6$
Now,
$2+5=5 \text { And } 2 \times 3=6$
Replacing the middle term $5 x$ by $2 x+3 x$, 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 34 Marks

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

Answer

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

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Question 44 Marks

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

Answer

The given expression is $2 x^2-3 x -2$
(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,
$(-4)+1=-3 \text { And }(-4) \times 1=-4$
Replacing the middle term 3 x by $-4 x + x$, 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 54 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, $(15)+(-8)=7$ 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 64 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 74 Marks

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

Answer

The given expression is $3 x^2+22 x+35$ (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., $3 \times 35=105$
Now, (15) $+(7)=22$ And $(15) \times(7)=105$
Replacing the middle term 22 x by 15 x 7 x ,
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 84 Marks

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

Answer

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

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Question 94 Marks

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

Answer

The given expression is $7 x^2-19 x-6$
(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., $7 \times(-6)=9$
Now,
$(-21)+2=19 \text { And }(-21) \times 2=-42$
Replacing the middle term $-19 x$ by $-21 x+x$, 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 104 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$) \times 3=-24$
Now,
$(-1)+24=23 \text { And }(-1) \times 24=-24$
Replacing the middle term 23 y by $- y +24 y$, 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 114 Marks

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

Answer

The given expression is $6 a^2+17 a b-3 b^2$
(co-effcient of $a^2=6$, co-effcient of $a=17 b$ and the constant term $=-3 b^2$ )
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., $6 \times\left(-3 b^2\right)=18 b^2$
Now,
$(18 b)+(- b )=17 b$ And $(18 b) \times(-b)=-18 b^2$
Replacing the middle term $17 a b$ by $-a b+18 a b$, 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 124 Marks

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

Answer

The given expression is $7 x-6-2 x^2$
(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$) \times(-6)=12$
Now,
$4+3=7 \text { And } 4 \times 3=12$
Replacing the middle term $7 x$ by $4 x+3 x$, 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 134 Marks

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

Answer

The given expression is $11 x^2-54 x+63$

(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 144 Marks

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

Answer

The given expression is $28-31 x-5 x^2$
(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 154 Marks

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

Answer

$a^2+2 a-3$
$= a ^2+2 a +\left(\frac{2}{2}\right)^2-\left(\frac{2}{2}\right)^2-3\left[\right.$ Adding and subtracting $\left(\frac{2}{2}\right)^2$, that is $\left.1^2\right]$
$=a 2+2 a+1-1-3$
$=(a+1) 2-4[$ 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 164 Marks

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

Answer

The given expression is $12 x^2-17 x y+6 y^2$
(Co-efficient of $x^2=12$, co-efficient of $x=17 y$ and the constant term $=6 y^2$ )
We will split the co-efficient of $x$ into two number parts such that their sum is $-17 y$ and their product equals to the product of the co-efficient of $x^2$ and constant term, i.e., $12 \times 6 y^2=72 y^2$
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 174 Marks

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

Answer

$4 x^2-12 x+5$
$=\Big(\text{x}^2 - 3\text{x} +\frac{5}{4}\Big)$ [Making the co-efficient of $\left.x^2=1\right]$
$=4\Big[\text{x}^2-3\text{x}+\Big(\frac{3}{2}\Big)^2-\Big(\frac{3}{2}\Big)^2+ \frac{5}{4}\Big]$ $\Big[$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 184 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 $p q=-54$
Now，
$6+(-9)=-3 \text { And } 6 \times(-9)=-54$
Splitting the middle term $-3 a$ in the given quadratic as $-9+6 a$, 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 194 Marks

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

Answer

The given expression is $6 x^2-5 x y+6 y^2$
(Co-efficient of $x^2=6$, co-efficient of $x=5 y$ and the constant term $=6 y^2$ )
We will split the co-efficient of $x$ into two number parts such that their sum is $-17 y$ and their product equals to the product of the co-efficient of $x^2$ and constant term, i.e., $6 \times\left(-6 y^2\right)=36 y^2$
Now,
$(-9 y)+(4 y)=-5 y \text { And }(-9 y) \times(4 y)=-36 y^2$
Replacing the middle term $-5 x y$ by $-9 x y+4 x y$, 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 204 Marks

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

Answer

The given expression is $36 a^2+12 a b c-15 b^2 c^2$
(co-effcient of $a^2=36, c o$-effcient of $a=12 b c$ and the constant term $=-15 b^2$ )
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., $36 \times\left(-15 b^2 c^2\right)=-540 b^2 c^2$
Now，
$(-18 bc)+(30 bc)=12 bc \text { And }(18 bc) \times(30 bc)=-540 b^2 c^2$
Replacing the middle term 17ab- by -ab + 18ab, we have:
$36 a^2+12 a b c-15 b^2 c^2=36 a^2-18 a b c+30 a b c-15 b^2 c^2$
$=\left(36 a^2-18 a b c\right)+\left(30 a b c-15 b^2 c^2\right)$
$=18 a(2 a-b c)+15 b c(2 a-b c)$
$=3(6 a+5 a b c)(2 a-b c)$

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Question 214 Marks

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

Answer

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

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Question 224 Marks

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

Answer

$4 y^2+12 y+54\left(y^2=3 y+\frac{5}{4}\right)\left[\right.$ making the co-efficient of $\left.y^2\right]$
$4\Big[\text{y}^2+3\text{y}+\Big(\frac{3}{2}\Big)^2-\Big(\frac{3}{2}\Big)^2=\frac{5}{4}$ $\Bigg[$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 234 Marks

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

Answer

The given expression is $14 x^2+11 x y-15 y^2$
(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. $\times\left(-15 y^2\right)=-210 y^2$
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:
$14 x^2+11 x y-15 y^2=14 x^2-10 x y+21 y-15 y^2$
$=\left(14 x^2-10 x y\right)+\left(21 x y-15 y^2\right)$
$=2 x(7 x-5 y)+3 y(7 x-5 y)$

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Question 244 Marks

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

Answer

The given expression is $6 x^2-13 x y+2 y^2$

(Co-efficient of $x^2=6$, co-efficient of $x=-13 y$ and the constant term $=2 y^2$ )

We will split the co-efficient of $x$ into two number parts such that their sum is $-13 y$ and their product equals to the product of the co-efficient of $x^2$ and constant term, i.e., $6 \times\left(2 y^2\right)=12 y^2$

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 254 Marks

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

Answer

$=z^2-4 z-12$
$\Big[$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 264 Marks

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

Answer

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

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