RS Aggarwal solution class 8 chapter 7 Factorisation Exercise 7A

Exercise 7A

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Q1 | Ex-7A | Class 8 | RS AGGARWAL | chapter 7 | Factorisation  | myhelper

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Question 1:

Factorise:
(i) 12x + 15
(ii) 14m − 21
(iii) 9n − 12n²

Answer 1:

 (i) $12x+15=3\left(4x+5\right)$
 (ii) $14m-21=7\left(2m-3\right)$
 (iii) $9n-12n^2=3n\left(3-4n\right)$

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Question 2:

Factorise:
(i) 16a² − 24ab
(ii) 15ab² − 20a²b
(iii) 12x²y³ − 21x³y²

Answer 2:

 (i) H.C.F. of $16a^2$ and $24ab$ is $8a$.

        ∴ $16a^2-24ab=8a\left(2a-3b\right)$
     
(ii) H.C.F. of $15a{b}^2$ and $20a^{2}b$ is $5ab$.

        ∴ $15a{b}^2-20a^{2}b=5ab\left(3b-4a\right)$

 (iii) ​H.C.F. of $12x^2{y}^3$ and $21x^3{y}^2$ is $3x^2{y}^2$.

        ∴ $12x^2{y}^3-21x^3{y}^2=3x^2{y}^2\left(4y-7x\right)$

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Question 3:

Factorise:
(i) 24x³ − 36x²y
(ii) 10x³ − 15x²
(iii) 36x³y − 60x²y³z

Answer 3:

(i) H.C.F. of $24x^3$ and $36x^{2}y$ is $12x^2$.

        ∴  $24x^3-36x^{2}y=12x^2\left(\mathrm{2x}-3y\right)$

(ii)  H.C.F. of $10x^3$ and $15x^2$ is $5x^3$.

         ∴ $10x^3-15x^2=5x^2\left(2x-3\right)$
   
(iii) H.C.F. of $36x^{3}y$ and $60x^2{y}^{3}z$ is $12x^{2}y$.

         ∴ $36x^{3}y-60x^2{y}^{3}z=12x^{2}y\left(3x-5y^{2}z\right)$

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Question 4:

Factorise:
(i) 9x³ − 6x² + 12x
(ii) 8x² − 72xy + 12x
(iii) 18a³b³ − 27a²b³ + 36a³b²

Answer 4:

(i) H.C.F. of $9x^3$, $6x^2$ and $12x$ is $3x$.

        ∴ $9x^3-6x^2+12x=3x\left(3x^2-2x+4\right)$

(ii) H.C.F. of $8x^3$, $72xy$ and $12x$ is $4x$.

       ∴  $8x^3-72xy+12x=4x\left(2x^2-18y+3\right)$

(iii) H.C.F. of $18a^3{b}^3$, $27a^2{b}^3$ and $36a^3{b}^2$ is $9a^2{b}^2$.

        ∴ $18a^3{b}^3-27a^2{b}^3+36a^3{b}^2=9a^2{b}^2\left(2ab-3b+4a\right)$

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Question 5:

Factorise:
(i) 14x³ + 21x⁴y − 28x²y²
(ii) −5 − 10t + 20t²

Answer 5:

(i) H.C.F. of $14x^3$, $21x^{4}y$ and $28x^2{y}^2$ is $7x^2$.

        ∴ $14x^3+21x^{4}y-28x^2{y}^2=7x^2\left(2x+3x^{2}y-4y^2\right)$

(ii) H.C.F. of $-5$, $-10t$ and $20t^2$ is 5.

         ∴ $-5-10t+20t^2=5\left(-1-2t+4t^2\right)$

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Question 6:

Factorise:
(i) x(x + 3) + 5(x + 3)
(ii) 5x(x − 4) − 7(x − 4)
(iii) 2m(1 − n) + 3(1 − n)

Answer 6:

(i) $x\left(x+3\right)+5\left(x+3\right)=\left(x+3\right)\left(x+5\right)$

(ii) $5x\left(x-4\right)-7\left(x-4\right)=\left(x-4\right)\left(5x-7\right)$

(iii) $2m\left(1-n\right)+3\left(1-n\right)=\left(1-n\right)\left(2m+3\right)$

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Question 7:

Factorise:
6a(a − 2b) + 5b(a − 2b)

Answer 7:

We have:
$6a\left(a-2b\right)+5b\left(a-2b\right)=\left(a-2b\right)\left(6a+5b\right)$

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Question 8:

Factorise:
x³(2a − b) + x²(2a − b)

Answer 8:

We have:
$x^3\left(2a-b\right)+x^2\left(2a-b\right)=\left(2a-b\right)\left(x^3+x^2\right)=x^2\left(x+1\right)\left(2a-b\right)$

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Question 9:

Factorise:
9a(3a − 5b) − 12a²(3a − 5b)

Answer 9:

We have:
$9a\left(3a-5b\right)-12a^2\left(3a-5b\right)=\left(3a-5b\right)\left(9a-12a^2\right)=3a\left(3a-5b\right)\left(3-4a\right)$

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Question 10:

Factorise:
(x + 5)² − 4(x + 5)

Answer 10:

We have:
${\left(x+5\right)}^2-4\left(x+5\right)=\left(x+5\right)\left\{\left(x+5\right)-4\right\}$
                         $$\begin{gathered} =\left(x+5\right)\left(x+5-4\right) \\ =\left(x+5\right)\left(x+1\right) \end{gathered}$$

∴ ${\left(x+5\right)}^2-4\left(x+5\right)=\left(x+5\right)\left(x+1\right)$
      

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Question 11:

Factorise:
3(a − 2b)² − 5(a − 2b)

Answer 11:

 $3{\left(a-2b\right)}^2-5\left(a-2b\right)=\left(a-2b\right)\left\{3\left(a-2b\right)-5\right\}$
                                    $=\left(a-2b\right)\left(3a-6b-5\right)$

∴ $3{\left(a-2b\right)}^2-5\left(a-2b\right)=\left(a-2b\right)\left(3a-6b-5\right)$

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Question 12:

Factorise:
2a + 6b − 3(a + 3b)²

Answer 12:

We have:
      $$\begin{gathered} 2a+6b-3{\left(a+3b\right)}^2=2\left(a+3b\right)-3{\left(a+3b\right)}^2 \\ =\left(a+3b\right)\left\{2-3\left(a+3b\right)\right\} \\ =\left(a+3b\right)\left(2-3a-9b\right) \end{gathered}$$

∴ $2a+6b-3{\left(a+3b\right)}^2=\left(a+3b\right)\left(2-3a-9b\right)$

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Question 13:

Factorise:
16(2p − 3q)² − 4(2p − 3q)

Answer 13:

 We have:
$$\begin{gathered} 16{\left(2p-3q\right)}^2-4\left(2p-3q\right)=\left(2p-3q\right)\left\{16\left(2p-3q\right)-4\right\} \\ =\left(2p-3q\right)\left(32p-48q-4\right) \end{gathered}$$

    ∴ $16{\left(2p-3q\right)}^2-4\left(2p-3q\right)=\left(2p-3q\right)\left(32p-48q-4\right)$

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Question 14:

Factorise:
x(a − 3) + y(3 − a)

Answer 14:

 We have:
$$\begin{gathered} x\left(a-3\right)+y\left(3-a\right)=x\left(a-3\right)-y\left(a-3\right) \\ =\left(a-3\right)\left(x-y\right) \end{gathered}$$

∴ $x\left(a-3\right)+y\left(3-a\right)=\left(a-3\right)\left(x-y\right)$

Q15 Ex-7A Class 8 RS AGGARWAL chapter 7 Factorisation

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Question 15:

Factorise:
12(2x − 3y)² − 16(3y − 2x)

Answer 15:

We have:
$$\begin{gathered} 12{\left(2x-3y\right)}^2-16\left(3y-2x\right)=12{\left(2x-3y\right)}^2+16\left(2x-3y\right) \\ =\left(2x-3y\right)\left\{12\left(2x-3y\right)+16\right\} \\ =\left(2x-3y\right)\left(24x-36y+16\right) \end{gathered}$$

$\therefore 12{\left(2x-3y\right)}^2-16\left(3y-2x\right)=\left(2x-3y\right)\left(24x-36y+16\right)$

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Question 16:

Factorise:
(x + y)(2x + 5) − (x + y)(x + 3)

Answer 16:

We have:
$$\begin{gathered} \left(x+y\right)\left(2x+5\right)-\left(x+y\right)\left(x+3\right)=\left(x+y\right)\left\{\left(2x+5\right)-\left(x+3\right)\right\} \\ =\left(x+y\right)\left(2x+5-x-3\right) \\ =\left(x+y\right)\left(x+2\right) \end{gathered}$$

Q17 Ex-7A Class 8 RS AGGARWAL chapter 7 Factorisation

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Question 17:

Factorise:
ar + br + at + bt

Answer 17:

By grouping the terms:
$$\begin{gathered} ar+br+at+bt=\left(ar+br\right)+\left(at+bt\right) \\ =r\left(a+b\right)+t\left(a+b\right) \\ =\left(a+b\right)\left(r+t\right) \end{gathered}$$

∴ $ar+br+at+bt=\left(a+b\right)\left(r+t\right)$

Q18 Ex-7A Class 8 RS AGGARWAL chapter 7 Factorisation

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Question 18:

Factorise:
x² − ax − bx + ab

Answer 18:

 By suitably arranging the terms:
         $$\begin{gathered} x^2-ax-bx+ab=x^2-bx-ax+ab \\ =\left(x^2-bx\right)-\left(ax-ab\right) \\ =x\left(x-b\right)-a\left(x-b\right) \\ =\left(x-b\right)\left(x-a\right) \end{gathered}$$

       ∴ $x^2-ax-bx+ab=\left(x-b\right)\left(x-a\right)$

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Question 19:

Factorise:
ab² − bc² − ab + c²

Answer 19:

 By suitably arranging the terms:
       $$\begin{gathered} a{b}^2-b{c}^2-ab+c^2=a{b}^2-ab-b{c}^2+c^2 \\ =\left(a{b}^2-ab\right)-\left(b{c}^2-c^2\right) \\ =ab\left(b-1\right)-c^2\left(b-1\right) \\ =\left(b-1\right)\left(ab-c^2\right) \end{gathered}$$

       ∴ $a{b}^2-b{c}^2-ab+c^2=\left(b-1\right)\left(ab-c^2\right)$

Q20 Ex-7A Class 8 RS AGGARWAL chapter 7 Factorisation

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Question 20:

Factorise:
x² − xz + xy − yz

Answer 20:

 By suitably arranging the terms:
      $$\begin{gathered} x^2-xz+xy-yz=x^2+xy-xz-yz \\ =\left(x^2+xy\right)-\left(xz+yz\right) \\ =x\left(x+y\right)-z\left(x+y\right) \\ =\left(x+y\right)\left(x-z\right) \end{gathered}$$ 

      ∴ $x^2-xz+xy-yz=\left(x+y\right)\left(x-z\right)$

Q21 Ex-7A Class 8 RS AGGARWAL chapter 7 Factorisation

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Question 21:

Factorise:
6ab − b² + 12ac − 2bc

Answer 21:

 By suitably arranging the terms:
      $$\begin{gathered} 6ab-b^2+12ac-2bc=6ab+12ac-b^2-2bc \\ =\left(6ab+12ac\right)-\left(b^2+2bc\right) \\ =6a\left(b+2c\right)-b\left(b+2c\right) \\ =\left(b+2c\right)\left(6a-b\right) \end{gathered}$$

     ∴ $6ab-b^2+12ac-2bc=\left(b+2c\right)\left(6a-b\right)$

Q22 Ex-7A Class 8 RS AGGARWAL chapter 7 Factorisation

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Question 22:

Factorise:
(x − 2y)² + 4x − 8y

Answer 22:

 We have:
       $$\begin{gathered} {\left(x-2y\right)}^2+4x-8y={\left(x-2y\right)}^2+4\left(x-2y\right) \\ =\left(x-2y\right)\left(x-2y\right)+4\left(x-2y\right) \\ =\left(x-2y\right)\left\{\left(x-2y\right)+4\right\} \\ =\left(x-2y\right)\left(x-2y+4\right) \end{gathered}$$

      ∴ ${\left(x-2y\right)}^2+4x-8y=\left(x-2y\right)\left(x-2y+4\right)$

Q23 Ex-7A Class 8 RS AGGARWAL chapter 7 Factorisation

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Question 23:

Factorise:
y² − xy(1 − x) − x³

Answer 23:

We have:
        $$\begin{gathered} y^2-xy\left(1-x\right)-x^3=y^2-xy+x^{2}y-x^3 \\ =\left(y^2-xy\right)+\left(x^{2}y-x^3\right) \\ =y\left(y-x\right)+x^2\left(y-x\right) \\ =\left(y-x\right)\left(y+x^2\right) \end{gathered}$$

       ∴ $y^2-xy\left(1-x\right)-x^3=\left(y-x\right)\left(y+x^2\right)$

Q24 Ex-7A Class 8 RS AGGARWAL chapter 7 Factorisation

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Question 24:

Factorise:
(ax + by)² + (bx − ay)²

Answer 24:

We have:
        $$\begin{gathered} {\left(ax+by\right)}^2+{\left(bx-ay\right)}^2=\left(a^2{x}^2+b^2{y}^2+2axby\right)+\left(b^2{x}^2+a^2{y}^2-2bxay\right) \\ =a^2{x}^2+a^2{y}^2+b^2{y}^2+b^2{x}^2+2axby-2bxay \\ =a^2\left(x^2+y^2\right)+b^2{x}^2+b^2{y}^2+2axby-2axby \\ =a^2\left(x^2+y^2\right)+b^2\left(x^2+y^2\right) \\ =\left(x^2+y^2\right)\left(a^2+b^2\right) \end{gathered}$$

       ∴ ${\left(ax+by\right)}^2+{\left(bx-ay\right)}^2=\left(x^2+y^2\right)\left(a^2+b^2\right)$

Q25 Ex-7A Class 8 RS AGGARWAL chapter 7 Factorisation

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Question 25:

Factorise:
ab² + (a − 1)b − 1

Answer 25:

We have:
       $$\begin{gathered} a{b}^2+\left(a-1\right)b-1=a{b}^2+ba-b-1 \\ =\left(a{b}^2+ba\right)-\left(b+1\right) \\ =ab\left(b+1\right)-1\left(b+1\right) \\ =\left(b+1\right)\left(ab-1\right) \end{gathered}$$

      ∴ $a{b}^2+\left(a-1\right)b-1=\left(b+1\right)\left(ab-1\right)$

Q26 Ex-7A Class 8 RS AGGARWAL chapter 7 Factorisation

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Question 26:

Factorise:
x³ − 3x² + x − 3

Answer 26:

We have:
        $$\begin{gathered} x^3-3x^2+x-3=\left(x^3-3x^2\right)+\left(x-3\right) \\ =x^2\left(x-3\right)+1\left(x-3\right) \\ =\left(x-3\right)\left(x^2+1\right) \end{gathered}$$

       ∴ $x^3-3x^2+x-3=\left(x-3\right)\left(x^2+1\right)$

Q27 Ex-7A Class 8 RS AGGARWAL chapter 7 Factorisation

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Question 27:

Factorise:
ab(x² + y²) − xy(a² + b²)

Answer 27:

We have:
      $$\begin{gathered} ab\left(x^2+y^2\right)-xy\left(a^2+b^2\right)=ab{x}^2+ab{y}^2-a^{2}xy-b^{2}xy \\ =ab{x}^2-a^{2}xy+ab{y}^2-b^{2}xy \\ =ax\left(bx-ay\right)+by\left(ay-bx\right) \\ =ax\left(bx-ay\right)-by\left(bx-ay\right) \\ =\left(bx-ay\right)\left(ax-by\right) \end{gathered}$$

      ∴ $ab\left(x^2+y^2\right)-xy\left(a^2+b^2\right)=\left(bx-ay\right)\left(ax-by\right)$

Q28 Ex-7A Class 8 RS AGGARWAL chapter 7 Factorisation

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Question 28:

Factorise:
x² − x(a + 2b) + 2ab

Answer 28:

We have:
        $x^2-x\left(a+2b\right)+2ab=x^2-ax-2bx+2ab$
                                       $$\begin{gathered} =x^2-2bx-ax+2ab \\ =\left(x^2-2bx\right)-\left(ax-2ab\right) \\ =x\left(x-2b\right)-a\left(x-2b\right) \\ =\left(x-2b\right)\left(x-a\right) \end{gathered}$$

      ∴ $x^2-x\left(a+2b\right)+2ab=\left(x-2b\right)\left(x-a\right)$