m(x-3y)^2+n(3y-x)+5x-15y factorise with grouping method
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1. Factor grouping the expressions:
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1 + a + ac + a2c
Solution:
1 + a + ac + a2c
= (1 + a) + (ac + a2c)
= (1 + a) + ac (1 + a)
= (1 + a) (1 + ac).
2. How to factor by grouping the following algebraic expressions?
(i) a2 - ac + ab - bc
Solution:
a2 - ac + ab - bc
= a(a - c) + b(a - c)
= (a - c) (a + b)
Therefore, by factoring expressions we get (a - c)(a + b)
(ii) a2 + 3a + ac + 3c
Solution:
a2 + 3a + ac + 3c
= a(a + 3) + c(a + 3)
= (a + 3) (a + c)
Therefore, by factoring expressions we get (a + 3)(a + c)
3. Factorize the algebraic expressions:
(i) 2x + cx + 2c + c2
Solution:
2x + cx + 2c + c2
= x(2 + c) + c(2 + c)
= (2 + c) (x + c)
(ii) x2 - ax + 5x - 5a
Solution:
x2 - ax + 5x - 5a
= x(x - a) + 5(x - a)
= (x - a) (x + 5)
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(iii) ax - bx - az + bz
Solution:
ax - bx - az + bz
= x(a - b) - z(a - b)
= (a - b) (x - z)
(iv) mx - 2my - nx + 2ny
Solution:
mx - 2my - nx + 2ny
= m(x - 2y) - n(x - 2y)
= (x - 2y) (m - n)
(v) ax2 - 3bxy – axy + 3by2
Solution:
ax2 - 3bxy – axy + 3by2
= x(ax – 3by) – y(ax – 3by)
= (ax - 3by) (ax - y)
4. Factor each of the following expressions by grouping:
(i) x2 - 3x - xy + 3y
Solution:
x2 - 3x - xy + 3y
= x(x – 3) – y(x – 3)
= (x – 3) (x – y)
(ii) ax2 + bx2 + 2a + 2b
Solution:
ax2 + bx2 + 2a + 2b
= x2(a + b) + 2(a + b)
= (a + b) (x2 + 2)
(iii) 2ax2 + 3axy - 2bxy - 3by2
Solution:
2ax2 + 3axy - 2bxy - 3by2
= ax(2x + 3y) - by(2x + 3y)
= (2x + 3y) (ax - by)
(iv) amx2 + bmxy – anxy – bny2
Solution:
amx2 + bmxy – anxy – bny2
= mx(ax + by) – ny(ax + by)
= (ax + by) (mx – ny)
5. Factorize:
(i) (x + y) (2x + 5) - (x + y) (x + 3)
Solution:
(x + y) (2x + 5) - (x + y) (x + 3)
= (x + y) [(2x + 5) - (x + 3)]
= (x + y) [2x + 5 - x -3]
= (x + y) (x + 2)
(ii) 6ab - b2 + 12ac - 2bc
Solution:
6ab - b2 + 12ac - 2bc
= b(6a - b) + 2c(6a - b)
= (6a - b) (b + 2c)
`
`
1 + a + ac + a2c
Solution:
1 + a + ac + a2c
= (1 + a) + (ac + a2c)
= (1 + a) + ac (1 + a)
= (1 + a) (1 + ac).
2. How to factor by grouping the following algebraic expressions?
(i) a2 - ac + ab - bc
Solution:
a2 - ac + ab - bc
= a(a - c) + b(a - c)
= (a - c) (a + b)
Therefore, by factoring expressions we get (a - c)(a + b)
(ii) a2 + 3a + ac + 3c
Solution:
a2 + 3a + ac + 3c
= a(a + 3) + c(a + 3)
= (a + 3) (a + c)
Therefore, by factoring expressions we get (a + 3)(a + c)
3. Factorize the algebraic expressions:
(i) 2x + cx + 2c + c2
Solution:
2x + cx + 2c + c2
= x(2 + c) + c(2 + c)
= (2 + c) (x + c)
(ii) x2 - ax + 5x - 5a
Solution:
x2 - ax + 5x - 5a
= x(x - a) + 5(x - a)
= (x - a) (x + 5)
`
(iii) ax - bx - az + bz
Solution:
ax - bx - az + bz
= x(a - b) - z(a - b)
= (a - b) (x - z)
(iv) mx - 2my - nx + 2ny
Solution:
mx - 2my - nx + 2ny
= m(x - 2y) - n(x - 2y)
= (x - 2y) (m - n)
(v) ax2 - 3bxy – axy + 3by2
Solution:
ax2 - 3bxy – axy + 3by2
= x(ax – 3by) – y(ax – 3by)
= (ax - 3by) (ax - y)
4. Factor each of the following expressions by grouping:
(i) x2 - 3x - xy + 3y
Solution:
x2 - 3x - xy + 3y
= x(x – 3) – y(x – 3)
= (x – 3) (x – y)
(ii) ax2 + bx2 + 2a + 2b
Solution:
ax2 + bx2 + 2a + 2b
= x2(a + b) + 2(a + b)
= (a + b) (x2 + 2)
(iii) 2ax2 + 3axy - 2bxy - 3by2
Solution:
2ax2 + 3axy - 2bxy - 3by2
= ax(2x + 3y) - by(2x + 3y)
= (2x + 3y) (ax - by)
(iv) amx2 + bmxy – anxy – bny2
Solution:
amx2 + bmxy – anxy – bny2
= mx(ax + by) – ny(ax + by)
= (ax + by) (mx – ny)
5. Factorize:
(i) (x + y) (2x + 5) - (x + y) (x + 3)
Solution:
(x + y) (2x + 5) - (x + y) (x + 3)
= (x + y) [(2x + 5) - (x + 3)]
= (x + y) [2x + 5 - x -3]
= (x + y) (x + 2)
(ii) 6ab - b2 + 12ac - 2bc
Solution:
6ab - b2 + 12ac - 2bc
= b(6a - b) + 2c(6a - b)
= (6a - b) (b + 2c)
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