# RD Sharma Solutions Class 8 Factorization Exercise 7.9

## RD Sharma Solutions Class 8 Chapter 7 Exercise 7.9

Solve: Q1. p2 + 6p + 8

Soln:

p2 + 6p + 8

= $p^{2} + 6p + \left ( \frac{6}{2} \right )^{2} – \left ( \frac{6}{2} \right )^{2} + 8 \; \; \; [Adding \; and \; subtracting \; \left ( \frac{6}{2} \right )^{2} , \; that \; is \; 3^{2} ]$

= p2 + 6p + 32 – 32 + 8

= p2 + 2 x p x 3 + 32 – 9 + 8

= p2 + 2 x p x 3 + 32 – 1

= (p + 3)2 – 12                           [Completing the square]

= [(p + 3) – 1][(p + 3) + 1]

= (p + 3 – 1) ( p + 3 + 1)

= (p + 2) (p + 4)

Q2. q2 – 10q + 21

Soln:

q2 – 10q + 21

= $q^{2} – 10q + \left ( \frac{10}{2} \right )^{2} – \left ( \frac{10}{2} \right )^{2} + 21 \; \; \; [Adding \; and \; subtracting \; \left ( \frac{10}{2} \right )^{2} , \; that \; is \; 5^{2} ]$

= q2 – 2 x q x 5 + 52 – 52 + 21

= (q – 5)2 – 4                        [Completing the square]

= [(q – 5) – 2] [(q – 5) + 2]

= (q – 5 – 2) (q – 5 + 2)

= (q – 7) (q – 3)

Q3. 4y2 + 12y + 5

Soln:

4y2 + 12y + 5

$\\ 4\left ( y^{2} + 3y + \frac{5}{4} \right ) \; \; \; \; [Making \; the \; co-efficient \; of \; y^{2}] \\ 4\left [ y^{2} + 3y + \left ( \frac{3}{2} \right )^{2} – \left (\frac{3}{2} \right )^{2} + \frac{5}{4}\right ] \; \; \; \left [Adding \; and \; subtracting \; \left ( \frac{3}{2} \right )^{2} \right ] \\ = 4 \left [ \left ( y + \frac{3}{2} \right )^{2} – \frac{9}{4} + \frac{5}{4} \right ] \\ = 4 \left [ \left ( y + \frac{3}{2} \right )^{2} – 1 \right ] \; \; \; [Completing \; the \; square] \\ = 4 \left [ \left ( y + \frac{3}{2} – 1 \right ) \right ] \left [ \left ( y + \frac{3}{2} + 1 \right ) \right ]$

$\\ = 4 \left ( y + \frac{3}{2} – 1 \right) \left ( y + \frac{3}{2} – 1\right ) \\ = 4 \left ( y + \frac{3}{2} \right) \left ( y + \frac{3}{2} \right ) \\ = (2y + 1)(2y + 5)$

Q4. p2 + 6p – 16

Soln:

p2 + 6p – 16

$\\ p^{2} + 6p + \left ( \frac{6}{2} \right )^{2} – \left ( \frac{6}{2} \right )^{2} – 16 \; \; \; \; \; \; \left [ Adding \; and \; subtracting \; \left ( \frac{6}{2} \right )^{2}, \; that \; is \; 3^{2} \right ]$

= p2 + 6p + 32 – 9 – 16

= (p + 3)2 – 25                       [Completing the square]

= (p + 3)2 – 52

= [(p + 3) – 5] [(p + 3) + 5]

= (p + 3 – 5) (p + 3 + 5)

= (p – 2)(p + 8)

Q5. x2 + 12x + 20

Soln:

x2 + 12x + 20

= $x^{2} + 12x + \left ( \frac{12}{2} \right )^{2} – \left ( \frac{12}{2} \right )^{2} + 20 \; \; \; \; \; \; \left [ Adding \; and \; subtracting \; \left ( \frac{12}{2} \right )^{2}, \; that \; is \; 6^{2} \right ]$

= x2 + 12x + 62 – 62 + 20

= (x + 6)2 – 16                         [completing the square]

= (x + 6)2 – 42

= [(x + 6) – 4] [(x + 6) + 4]

= (x + 6 – 4) (x + 6 + 4)

= (x + 2) (x + 10)

Q6. a2 – 14a – 51

Soln:

a2 – 14a – 51

= $a^{2} – 14a + \left ( \frac{14}{2} \right )^{2} – \left ( \frac{14}{2} \right )^{2} – 51 \; \; \; \; \; \; \left [ Adding \; and \; subtracting \; \left ( \frac{14}{2} \right )^{2}, \; that \; is \; 7^{2} \right ]$

= a2 – 14a + 7– 72 – 51

= (a – 7)2 – 100                   [Completing the square]

= (a – 7)2 – 102

= [(a – 7) – 10] [(a – 7) + 10]

= (a – 7 – 10) (a – 7 + 10)

= (a – 17)(a + 3)

Q7. a2 + 2a – 3

Soln:

a2 + 2a – 3

= $a^{2} + 2a + \left ( \frac{2}{2} \right )^{2} – \left ( \frac{2}{2} \right )^{2} – 3 \; \; \; \; \; \; \left [ Adding \; and \; subtracting \; \left ( \frac{2}{2} \right )^{2}, \; that \; is \; 1^{2} \right ]$

= a2 + 2a + 1 – 1 – 3

= (a + 1)2 – 4                                       [Completing the square]

= (a + 1)2 – 22

= [(a + 1) – 2] [(a + 1) + 2]

= (a + 1 – 2) (a + 1 + 2)

= (a – 1)(a + 3)

Q8. 4x2 – 12x + 5

Soln:

4x2 – 12x + 5

= $4 (x^{2} – 3x + \frac{5}{4})$                [Making the co-efficient of x2 = 1]

= $\\ 4\left [ x^{2} – 3x + \left ( \frac{3}{2} \right )^{2} – \left ( \frac{3}{2} \right )^{2} + \frac{5}{4} \right ] \; \; \; \; \left [ Adding \; and \; subtracting \left ( \frac{3}{2} \right )^{2} \right ]\\ = 4\left [ \left ( x – \frac{3}{2} \right )^{2} – \frac{9}{4} + \frac{5}{4}\right ] [Completing \; the \; square] \\ = 4\left [ \left ( x – \frac{3}{2} \right )^{2} – 1 \right ] \\ = 4\left [ \left ( x – \frac{3}{2} \right ) – 1 \right ] \left [ \left ( x – \frac{3}{2} \right ) + 1 \right ] \\ = 4 \left ( x – \frac{3}{2} – 1 \right ) \left ( x – \frac{3}{2} + 1 \right ) \\ = 4 \left ( x – \frac{5}{2} \right ) \left ( x – \frac{1}{2} \right ) \\ = (2x – 5)(2x – 1)$

Q9. (y – 3)(y – 4)

Soln:

= y2-7y+12

(Adding and subtracting $(\frac{7}{2})^{2}$)

= y2-7y+$(\frac{7}{2})^{2}$$(\frac{7}{2})^{2}$+12

Completing the square

= (y-$(\frac{7}{2})$)2$\frac{49}{4}$+ $\frac{48}{4}$

= (y-$(\frac{7}{2})$)2$(\frac{1}{4})$

= (y-$(\frac{7}{2}$)2 –($\frac{1}{2}^{2}$)

= [(y-$(\frac{7}{2}$$\frac{1}{2}$) [(y-$(\frac{7}{2}$+$\frac{1}{2}$)

= [(y-$(\frac{7}{2}$$\frac{1}{2}$)( [(y-$(\frac{7}{2}$+$\frac{1}{2}$)

= (y-4)(y-3)

Q10. (z – 6)(z + 2)

Soln:

= z2-4z-12

(Adding and subtracting $(\frac{4}{2})^{2}$)

= z2-4z+$(\frac{4}{2})^{2}$$(\frac{4}{2})^{2}$–12

= z2-4z+(2)2-(2)2-12

= (z-2)2 – 16

Completing the squares

= (z-2)2-(4)2

=[(z-2)-4][(z-2)+4]

= (z-6)(z+2)

#### Practise This Question

Sonali's best friends name started with the letter 'H'. She was working out the lines of symmetry in 'H'. Which of the following statements for letter H is/are incorrect.
Statement 1: It has only a horizontal line of symmetry.
Statement 2: It has only a vertical line of symmetry.
Statement 3: It has both a horizontal and a vertical line of symmetry.

Statement 4: It is asymmetrical.