# BSE Odisha 8th Class Maths Solutions Algebra Chapter 4 ଉତ୍ପାଦକୀକରଣ Ex 4(b)

Odisha State Board BSE Odisha 8th Class Maths Solutions Algebra Chapter 4 ଉତ୍ପାଦକୀକରଣ Ex 4(b) Textbook Exercise Questions and Answers.

## BSE Odisha Class 8 Maths Solutions Algebra Chapter 4 ଉତ୍ପାଦକୀକରଣ Ex 4(b)

Question 1.
x² + xy + 8x + 8y
ଡ –
x² + xy + 8x + 8yx(x + y) + 8(x + y)(x + y)(x + 8)

Question 2.
pq + pr + q² + qr
ଡ –
pq + pr + q² + qr = p(q + r) + q(q + r) = (q + r)(p + q)

Question 3.
ab+db+ac+dc
ଡ –
ab + db + ac + dc = b(a + d) + c(a + d) = (a+d)(b+c)

Question 4.
pq + qr + pr + r²
ଡ –
pq + qr + pr + r² = q(p + r) + r(p + r) = (p + r)(q + r)

Question 5.
15xy – 6x + 5y – 2
ଡ –
15xy – 6x + 5y – 2 = 3x(5y – 2)+ 1(5y – 2) = (5y – 2)(3x + 1)

Question 6.
ax + bx – ay – by
ଡ –
ax + bx – ay – by = x(a + b) – y(a + b) = (a + b)(x – y)

Question 7.
15pq + 15 + 9q + 25p
ଡ –
15Pq + 15 + 9q + 25p = 15pq + 9q + 25p + 15
= 3q(5p + 3) + 5(5p + 3) = (5p + 3)(3q + 5)

Question 8.
2a + 6b – 3(a + 3b)²
ଡ –
2a + 6b – 3(a + 3b)² = 2(a + 3b) – 3(a + 3b)(a + 3b)

Question 9.
a² + 2a + ab + 2b
ଡ –
a² + 2a + ab + 2b = a(a + 2) + b(a + 2) = (a + 2)(a + b)

Question 10.
x² – xz + xy – yz
ଡ –
x² – xz + xy – yz = x(x – z) + y(x – z) = (x – z)(x + y)

Question 11.
a² + bc – ba – ac
ଡ –
a² + bc – ba – ac = a² – ba – ac + bc = a(a – b) – c(a – b) = (a – b) (a – c)

Question 12.
2p² – pq – 2pr + qr
ଡ –
2p² – q – 2pr + qr = p(2p – q) -r²(p – q) = (2p – q)(p – r²)

Question 13.
x² – 3x + 2x – 6
ଡ –
x² – 3x + 2x – 6 = x(x – 3) + 2(x – 3) = (x – 3)(x + 2)

Question 14.
2x² – 5x + 4x – 10
ଡ –
2x² – 5x + 4x – 10 = x(2x – 5) + 2(2x – 5) = (2x – 5)(x + 2)

Question 15.
x² – y² + x – xy²
ଡ –
x² – y² + x – xy² = x² + x – xy² – y² = x(x + 1) – y²(x + 1) = (x + 1)(x – y²)

Question 16.
lm² – mn² – lm + n²
ଡ –
lm² – mn² – lm + n² = lm² – lm – mn² + n²
= lm(m – 1) – n(m – 1) = (m – 1) (lm – n²)

Question 17.
x³ – 2x²y + 3xy² – 6y³
ଡ –
x³ – 2x²y + 3xy² – 6y³ = x² (x – 2y) + 3y² (x – 2y) = (x – 2y) (x² + 3y²)

Question 18.
6ab – b² + 12ac – 2bc
ଡ –
6ab – b² + 12ac – 2bc = 6ab + 12ac – b² – 2bc
=6a(b + 2c) – b(b + 2c) = (b + 2c)(6a – b)

Question 19.
x² – 11xy – x + 11y
ଡ –
x² – 11xy – x + 11y = x(x – 11y) – 1(x – 11y) = (x – 11y)(x – 1)

Question 20.
3ax – 6ay – 85y + 4bx
ଡ –
3ax – 6ay – 8by + 4bx = 3ax – 6ay + 4bx – 8by
= 3a (x – 2y) + 4b(x – 2y) = (x – 2y) (3a + 4b)