Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 3.6 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 3 Polynomials.

## Practice Set 3.6 Algebra 9th Std Maths Part 1 Answers Chapter 3 Polynomials

Question 1.

Find the factors of the polynomials given below:

i. 2x^{2} + x – 1

ii. 2m^{2} + 5m – 3

iii. 12x^{2} + 61x + 77

iv. 3y^{2} – 2y – 1

v. √3x^{2} + 4x + √3

vi. \(\frac { 1 }{ 2 }\)x^{2} – 3x + 4

Solution:

i. 2x^{2} + x – 1

= 2x^{2} + 2x – x – 1

= 2x(x + 1)- 1(x + 1)

= (x + 1)(2x – 1)

ii. 2m^{2} + 5m – 3

= 2m^{2} + 6m – m – 3

= 2m(m + 3) – 1(m + 3)

= (m + 3)(2m – 1)

iii. 12x^{2} + 61x + 77

= 12x^{2} + 28x + 33x + 77

= 4x(3x + 7) 4 + 11(3x + 7)

= (3x + 7)(4x + 11)

iv. 3y^{2} – 2y – 1

= 3y^{2} – 3y + y – 1

= 3y(y – 1) + 1 (y – 1)

= (y – 1)(3y + 1)

v. √3×2 + 4x + √3

= √3×2 + 3x + x + √3

= √3×2 + √3 x √3x + x + √3

= √3x(x + √3) + 1 ( x + √3 )

= (x + √3)(√3x + 1)

vi. \(\frac { 1 }{ 2 }\) x^{2} – 3x + 4

= \(\frac { 1 }{ 2 }\) x^{2} – 2x – x + 4

= \(\frac{1}{2} x^{2}-\frac{2 \times 2}{2} x-x+4\)

= \(\frac { 1 }{ 2 }\) x(x – 4) – 1 (x – 4)

= (x – 4) (\(\frac { 1 }{ 2 }\) x – 1)

Alternate method

\(\frac { 1 }{ 2 }\) x^{2} – 3x + 4 = \(\frac { 1 }{ 2 }\) (x^{2} – 6x + 8)

= \(\frac { 1 }{ 2 }\) (x^{2} – 4x – 2x + 8)

= \(\frac { 1 }{ 2 }\) [x(x – 4) – 2(x – 4)]

= \(\frac { 1 }{ 2 }\) (x – 2)(x – 4)

Question 2.

Factorize the following polynomials.

i. (x^{2} – x)^{2} – 8(x^{2} – x) + 12

iii. (x^{2} – 6x)^{2} – 8(x^{2} – 6x + 8) – 64

v. (y + 2) (y – 3) (y + 8) (y + 3) + 56

vii. (x – 3) (x – 4)^{2} (x – 5) – 6

Solution:

i. (x^{2} – x)^{2} – 8(x^{2} – x) + 12

= m^{2} – 8m + 12 …[Putting x^{2} – x = m]

= m^{2} – 6m – 2m + 12

= m(m – 6) – 2(m – 6)

= (m – 6)(m – 2)

= (x^{2} – x- 6) (x^{2} – x- 2) …[Replacing m = x^{2} -x]

= (x^{2} – 3x + 2x – 6) (x^{2} – 2x + x – 2)

= [x(x – 3) + 2(x – 3)] [x(x – 2) + 1 (x-2)]

= (x – 3) (x + 2) (x – 2) (x + 1)

ii. (x – 5)^{2} – (5x – 25) – 24

= (x – 5)^{2} – (5x – 25) – 24

= (x – 5)^{2} – 5(x – 5) – 24

= m^{2} – 5m – 24 … [Putting x – 5 = m]

= m^{2} – 8m + 3m – 24

= m(m – 8) + 3(m – 8)

= (m – 8) (m + 3)

= (x – 5 – 8) (x – 5 + 3) … [Replacing m = x – 5]

= (x – 13) (x – 2)

iii. (x^{2} – 6x)^{2} – 8(x^{2} – 6x + 8) – 64

= m^{2} – 8(m + 8)-64 …[Putting x^{2} – 6x = m]

= m^{2} – 8m – 64 – 64

= m^{2} – 8m – 128

= m^{2} – 16m + 8m- 128

= m(m – 16) + 8(m – 16)

= (m – 16)(m + 8)

= (x^{2} – 6x – 16) (x^{2} – 6x + 8) … [Replacing m = x^{2} – 6x]

= (x^{2} – 8x + 2x – 16) (x^{2} – 4x – 2x + 8)

= [x(x – 8) + 2(x – 8)] [x(x – 4) – 2(x – 4)]

= (x – 8) (x + 2) (x – 4) (x – 2)

iv. (x^{2}– 2x + 3) (x^{2} – 2x + 5) – 35

= (m + 3) (m + 5) – 35 … [Putting x^{2} – 2x = m]

= m (m + 5) + 3(m + 5) – 35

= m^{2} + 5m + 3m + 15 – 35

= m^{2} + 8m – 20

= m^{2} + 10m – 2m – 20

= m(m + 10) – 2(m + 10)

= (m + 10) (m – 2)

= (x^{2} – 2x + 10) (x^{2} – 2x – 2) … [Replacing m = x^{2} – 2x]

v. (y + 2) (y – 3) (y + 8) (y + 3) + 56

= (y + 2)(y + 3)(y – 3)(y + 8) + 56

= (y^{2} + 3y + 2y + 6) (y^{2} + 8y – 3y – 24) + 56

= (y^{2} + 5y + 6) (y^{2} + 5y – 24) + 56

= (m + 6) (m – 24) + 56 … [Putting y2 + 5y = m]

= m (m – 24) + 6 (m – 24) + 56

= m^{2} – 24m + 6m – 144 + 56

= m^{2} – 18m – 88

= m^{2} – 22m + 4m – 88

= m(m – 22) + 4(m – 22)

= (m – 22) (m + 4)

= (y^{2} + 5y – 22)(y^{2} + 5y + 4) … [Replacing m = y^{2} + 5y]

= (y^{2} + 5y – 22) (y^{2} + 4y + y + 4)

= (y^{2} + 5y – 22) [y(y + 4) + 1(y + 4)]

= (y^{2} + 5y – 22) (y + 4) (y + 1)

vi. (y^{2} + 5y) (y^{2} + 5y – 2) – 24

= (m)(m – 2) – 24 … [Putting y^{2} + 5y = m]

= m^{2} – 2m – 24

= m^{2} – 6m + 4m – 24

= m(m – 6) + 4(m – 6)

= (m – 6) (m + 4)

= (y^{2} + 5y – 6) (y^{2} + 5y + 4) … [Replacing m = y^{2} + 5y]

= (y^{2} + 6y – y – 6) (y^{2} + 4y + y + 4)

= [y(y + 6) – 1(y + 6)] [y(y + 4) + 1(y + 4)]

= (y + 6) (y – 1) (y + 4) (y + 1)

vii. (x – 3) (x – 4)2 (x – 5) – 6

= (x – 3) (x – 5) (x – 4)^{2} – 6

= (x^{2} – 5x – 3x + 15) (x^{2} – 8x + 16) – 6

= (x^{2} – 8x + 15) (x^{2} – 8x + 16) – 6

= (m + 15) (m+ 16) – 6 … [Putting x^{2} – 8x = m]

= m (m + 16) + 15 (m + 16) – 6

= m^{2} + 16m + 15m + 240 – 6

= m^{2} + 31m + 234

= m^{2} + 18m + 13m + 234

= m(m + 18) + 13(m + 18)

= (m + 18) (m + 13)

= (x^{2} – 8x + 18) (x^{2} – 8x + 13) … [Replacing m = x^{2} – 8x]