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Solved Examples on Factoring by Grouping
1. Factor the following
(i) 12p + 15
(ii) 14p – 21
(iii) 9q – 12q²
Solution:
(i) The given expression is 12p + 15.
Expand the given expression. That is,
(3 * 4)p + (3 * 5).
Factor out the greatest common factor. That is,
3(4p + 5).
Then, 12p + 15 is equal to 3(4p + 5).
(ii) The given expression is 14p – 21.
Expand the given expression. That is,
(7 * 2)p – (7 * 3).
Factor out the greatest common factor. That is,
7(2p – 3).
Then, 14p – 21 is equal to 7(2p – 3).
(iii) The given expression is 9q – 12q^2.
Expand the given expression. That is,
(3 * 3)q – (3 * 4)q^2.
Factor out the greatest common factor. That is,
3q(3 – 4q).
Then, 9q – 12q^2 is equal to 3q(3 – 4q).
2. Factor by grouping the expressions
(i) 16x² – 24xy
(ii) 15xy² – 20x²y
(iii) 12a²b³ – 21a³b²
Solution:
(i) The given expression is 16x^2 – 24xy.
Expand the given expression. That is,
(4 * 4)x^2 – (4 * 6)xy.
Factor out the greatest common factor. That is,
4x(4x – 6y).
Then, 16x^2 – 24xy is equal to 4x(4x – 6y).
(ii) The given expression is 15xy^2 – 20x^2y.
Expand the given expression. That is,
(5 * 3)xy^2 – (5 * 4)x^2y.
Factor out the greatest common factor. That is,
5xy(3y – 4x).
Then, 15xy^2 – 20x^2y is equal to 5xy(3y – 4x).
(iii) The given expression is 12a^2b^3 – 21a^3b^2.
Expand the given expression. That is,
(3 * 4)a^2b^3 – (3 * 7)a^3b^2.
Factor out the greatest common factor. That is,
3a^2b^2(4b – 7a).
Then, 12a^2b^3 – 21a^3b^2 is equal to 3a^2b^2(4b – 7a).
3. Factorize the expressions
(i) 24a³ – 36a²b.
(ii) 10a³ – 15a²
(iii) 36a³b – 60a²b³c
Solution:
(i) The given expression is 24a^3 – 36a^2b.
Expand the given expression. That is,
(6 * 4)a^3 – (6 * 6)a^2b.
Factor out the greatest common factor. That is,
6a^2(4a – 6b).
Then, 24a^3 – 36a^2b is equal to 6a^2(4a – 6b).
(ii) The given expression is 10a^3 – 15a^2.
Expand the given expression. That is,
(5 * 2)a^3 – (5 * 3)a^2.
Factor out the greatest common factor. That is,
5a^2(2a – 3).
Then, 10a^3 – 15a^2 is equal to 5a^2(2a – 3).
(iii) The given expression is 36a^3b – 60a^2b^3c.
Expand the given expression. That is,
(6 * 6)a^3b – (6 * 10)a^2b^3c.
Factor out the greatest common factor. That is,
6a^2b(6a – 10b^2c).
Then, 36a^3b – 60a^2b^3c is equal to 6a^2b(6a – 10b^2c).
4. Factorize
(i) 9a³ – 6a² + 12a.
(ii) 8a² – 72ab + 12a.
(iii) 18x³y³ – 27x²y³ + 36x³y²
Solution:
(i) The given expression is 9a³ – 6a² + 12a.
Expand the given expression. That is,
(3 * 3)a^3 – (3 * 2)a^2 + (3 * 4)a.
Factor out the greatest common factor. That is,
3a(3a^2 – 2a + 4).
Then, 9a^3 – 6a^2 + 12a is equal to 3a(3a^2 – 2a + 4).
(ii) The given expression is 8a^2 – 72ab + 12a.
Expand the given expression. That is,
(4 * 2)a^2 – (4 * 18)ab + (4 * 3)a.
Factor out the greatest common factor. That is,
4a(2a – 18b + 3a).
Then, 8a^2 – 72ab + 12a is equal to 4a(2a – 18b + 3a).
(iii) The given expression is 18x³y³ – 27x²y³ + 36x³y²
Expand the given expression. That is,
(9 * 2)x^3y^3 – (9 * 3)x^2y^3 + (9 * 4)x^3y^2.
Factor out the greatest common factor. That is,
9x^2y^2(2xy -3y + 4x).
Then, 18x³y³ – 27x²y³ + 36x³y² is equal to 9x^2y^2 (2xy -3y + 4x).
5. How to factor by grouping?
(i) 14a³ + 21a^4b – 28a²b²
(ii) -5 – 10x + 20x²
Solution:
(i) The given expression is 14a^3 + 21a^4b – 28a^2b^2.
Expand the given expression. That is,
(7 * 2)a^3 + (7 * 3)a^4b – (7 * 4)a^2b^2.
Factor out the greatest common factor. That is,
7a^2(2a + 3a^2b – 4b^2).
Then, 14a^3 + 21a^4b – 28a^2b^2 is equal to 7a^2(2a + 3a^2b – 4b^2).
(ii) The given expression is -5 – 10x + 20x^2.
Expand the given expression. That is,
-5 – (5 * 2)x + (5 * 4)x^2.
Factor out the greatest common factor. That is,
5( -1 – 2x + 4x^2).
Then, -5 – 10x + 20x^2 is equal to 5( -1 – 2x + 4x^2) .
6. Factoring
(i) a (a + 3) + 5(a + 3)
(ii) 5a (a – 4) – 7 (a – 4).
(iii) 2x (1 – y) + 3(1 – y).
Solution:
(i) The given expression is a(a + 3) + 5(a + 3).
Factor out the greatest common factor. That is,
(a + 3) (a + 5)
Then,a (a + 3) + 5(a + 3) is equal to (a + 3) (a + 5).
(ii) The given expression is 5a(a – 4) – 7(a – 4).
Factor out the greatest common factor. That is,
(a – 4) (5a – 7).
Then, 5a (a – 4) – 7 (a – 4) is equal to (a – 4) (5a – 7).
(iii) The given expression is 2x(1 – y) + 3(1 – y).
Factor out the greatest common factor. That is,
(1 – y) (2x + 3).
Then, 2x(1 – y) + 3(1- y) is equal to (1 – y) (2x + 3).
7. Factoring the expressions
(i) 6x (x – 2y) + 5y (x – 2y).
(ii) p³ (2x – y) + p²(2x – y).
Solution:
(i) The given expression is 6x (x – 2y) + 5y (x – 2y).
Factor out the greatest common factor. That is,
(x – 2y) (6x + 5y).
Then, 6x (x – 2y) + 5y (x – 2y) is equal to (x – 2y) (6x + 5y).
(ii) The given expression is p^3 (2x – y) + p^2 (2x – y).
Factor out the greatest common factor. That is,
(2x – y)p^2 (p + 1).
Then, p^3 (2x – y) + p^2 (2x – y) is equal to p^2(2x – y) (p + 1).
8. How to factor by grouping polynomials?
(i) 9x (3x – 5y) – 12x²(3x – 5y)
(ii) (a + 5)² – 4 (a + 5)
(iii) 3(x – 2y)² – 5(x – 2y)
Solution:
(i) The given expression is 9x (3x – 5y) – 12x²(3x – 5y).
Factor out the greatest common factor. That is,
3x(3x – 5y)(3 – 4x).
Then, 9x (3x – 5y) – 12x²(3x – 5y) is equal to 3x(3x – 5y)(3 – 4x).
(ii) The given expression is (a + 5)^2 – 4 (a + 5).
Factor out the greatest common factor. That is,
(a + 5) [(a + 5) – 4] = (a + 5) (a + 1).
Then, (a + 5)^2 – 4 (a + 5) is equal to (a + 5) (a + 1).
(iii) The given expression is 3(x – 2y)² – 5(x – 2y).
Factor out the greatest common factor. That is,
(x – 2y) [3(x – 2y) – 5].
Then, 3(x – 2y)² – 5(x – 2y) is equal to (x – 2y) [3(x – 2y) – 5].
9. Factor completely
(i) 2x + 6y – 3(x + 3y)²
(ii) 16(2x – 3y)² – 4(2x – 3y)
(iii) p (x – 3) + q (3 – x)
(iv) 12 (2a – 3b)² – 16 (3b – 2a).
(v) (a + b)(2a + 5) – (a + b)(a + 3).
Solution:
(i) The given expression is 2x + 6y – 3(x + 3y)^2.
Factor out the greatest common factor. That is,
Factor out the greatest common factor. That is,
2(x + 3y) – 3(x + 3y)^2 = (x + 3y) (2 – 3(x + 3y)).
Then, 2x + 6y – 3(x + 3y)^2 is equal to (x + 3y) (2 – 3(x + 3y)).
(ii) The given expression is 16(2x – 3y)² – 4(2x – 3y).
Factor out the greatest common factor. That is,
4(2x – 3y) [4(2x – 3y) – 1].
Then, 16(2x – 3y)² – 4(2x – 3y) is equal to 4(2x – 3y) [4(2x – 3y) – 1].
(iii) The given expression is
p (x – 3) + q (3 – x).
we can write it as p (x – 3) – q(x – 3).
Factor out the greatest common factor. That is,
(x – 3) (p – q).
Then, p (x – 3) + q (3 – x) is equal to (x – 3) (p – q).
(iv) The given expression is 12 (2a – 3b)² – 16 (3b – 2a).
Expand the above expression. That is,
12 (2a – 3b)^2 + 32a – 48b.
We can write it as 12 (2a – 3b)^2 + 16 (2a – 3b).
Factor out the greatest common factor. That is,
4(2a – 3b) [3(2a – 3b) + 4].
Then, 12 (2a – 3b)² – 16 (3b – 2a) is equal to 4(2a – 3b) [3(2a – 3b) + 4].
(v) The given expression is (a + b)(2a + 5) – (a + b)(a + 3).
Factor out the greatest common factor. That is,
(a + b) [(2a + 5) – (a + 3)] = (a + b) (a + 2).
Then, (a + b)(2a + 5) – (a + b)(a + 3) is equal to (a + b) (a + 2).
10. How to factor by grouping polynomials?
(i) xp + yp + xq + yq.
(ii) p² – sp – qp + sq.
(iii) xy² – yz² – xy + z²
(iv) a² – ac + ab–bc.
(v) 6xy – y² + 12xz – 2yz.
Solution:
(i) The given expression is xp + yp + xq + yq.
Factor out the greatest common factor. That is,
p(x + y) + q(x + y) = (x + y) (p + q).
Then, xp + yp + xq + yq is equal to (x + y) (p + q).
(ii) The given expression is p2 – sp – qp + sq.
Factor out the greatest common factor. That is,
p(p – s) – q(p – s) = (p – s) (p – q).
Then, p2 – sp – qp + sq is equal to (p – s) (p – q).
(iii) The given expression is xy² – yz² – xy + z²
Factor out the greatest common factor. That is,
xy^2 – xy – yz^2 + z^2 = xy(y – 1) – z(yz – z).
Then, xy² – yz² – xy + z² is equal to xy(y – 1) – z(yz – z).
(iv) : The given expression is a2 – ac + ab – bc.
Factor out the greatest common factor. That is,
a(a – c) + b(a – c) = (a – c) (a + b).
Then, a2 – ac + ab – bc is equal to (a – c) (a + b).
(v) The given expression is 6xy – y² + 12xz – 2yz.
Factor out the greatest common factor. That is,
y(6x – y) + 2z(6x – y) = (6x – y) (y + 2z).
Then, 6xy – y² + 12xz – 2yz is equal to (6x – y) (y + 2z).
11. Solve by factoring
(i) (a – 2b)² + 4a–8b
(ii) b² – ab (1 – a) – a³
(iii) (rp + sq)² + (sp – rq)²
(iv) pq² + (p – 1)q – 1
Solution:
(i) The given expression is (a – 2b)² + 4a–8b.
Factor out the greatest common factor. That is,
(a – 2b)² + 4(a – 2b) = (a – 2b) (a – 2b + 4).
Then, (a – 2b)² + 4a–8b is equal to (a – 2b) (a – 2b + 4).
(ii) The given expression is b² – ab (1 – a) – a³
we can write it as b^2 – ab + a^2b – a^3.
Factor out the greatest common factor. That is,
b(b – a) + a^2(b – a) = (b – a)(b + a^2).
Then, b² – ab (1 – a) – a³ is equal to (b – a)(b + a^2).
(iii) The given expression is (rp + sq)² + (sp – rq)²
We can write it as (rp)^2 + (sq)^2 + 2rpsq + (sp)^2 + (rq)^2 – 2rpsq
it is equal to (rp)^2 + (sq)^2 + (sp)^2 + (rq)^2
Factor out the greatest common factor. That is,
r^2(p^2 + q^2) + s^2(p^2 + q^2) = (p^2 + q^2) (r^2 + s^2)
Then, (rp + sq)² + (sp – rq)² is equal to (p^2 + q^2) (r^2 + s^2)
(iv) The given expression is pq² + (p – 1)q – 1.
we can write it as pq^2 + pq – q – 1= pq(q +1) – (q + 1)
Factor out the greatest common factor. That is,
(q + 1) (pq – 1).
Then, pq² + (p – 1)q – 1 is equal to (q + 1) (pq – 1).
12. Factoring Algebraic Expressions
(i) p³ – 3p² + p – 3
(ii) xy (p² + q²) – pq (x² + y²)
(iii) p² – p(x + 2y) + 2xy
Solution:
(i) The given expression is p3 – 3p2 + p – 3.
We can write it as p^2(p – 3) + (p – 3)
Factor out the greatest common factor. That is, (p – 3) (p^2 + 1).
Then, p3 – 3p2 + p – 3is equal to (p – 3) (p^2 + 1).
(ii) The given expression is xy (p2 + q2) – pq (x2 + y2).
We can write it as xyp^2 + xyq^2 – pqx^2 – pqy^2.
Factor out the greatest common factor. That is, xp (yp –qx) – yq (yp–qx) = (yp – qx) (xp – yq).
Then, xy (p2 + q2) – pq (x2 + y2)is equal to (yp – qx) (xp – yq).
(iii) The given expression is p2 – p(x + 2y) + 2xy.
We can write it as p^2 – px – 2py + 2xy.
Factor out the greatest common factor. That is, p(p -2y) – x(p – 2y) = (p – 2y) (p – x).
Then, p2 – p(x + 2y) + 2xy is equal to (p – 2y) (p – x).