Answer :
To factorise each given polynomial completely, we shall follow the general strategy of looking for common factors and then applying algebraic identities or basic factorisation techniques. Let's tackle each part step-by-step:
### a. Factorise [tex]\(x^2 - 8xy\)[/tex]
First, we look for a common factor in both terms:
[tex]\[ x^2 - 8xy \][/tex]
Both terms have a common factor of [tex]\(x\)[/tex]:
[tex]\[ x(x - 8y) \][/tex]
Thus, the factorised form is:
[tex]\[ \boxed{x(x - 8y)} \][/tex]
### b. Factorise [tex]\(6q^3 - 2q + 2q^4\)[/tex]
First, we arrange the terms in standard form:
[tex]\[ 2q^4 + 6q^3 - 2q \][/tex]
Observe the common factor in all terms, which is [tex]\(2q\)[/tex]:
[tex]\[ 2q(q^3 + 3q^2 - 1) \][/tex]
So, the factorised form is:
[tex]\[ \boxed{2q(q^3 + 3q^2 - 1)} \][/tex]
### c. Factorise [tex]\(7mp^3 + 21m^2p^2 - 14mp^4\)[/tex]
First, identify the common factor:
[tex]\[ 7mp^3 + 21m^2p^2 - 14mp^4 \][/tex]
Factor out [tex]\(7m\)[/tex]:
[tex]\[ 7m(p^3 + 3m p^2 - 2p^4) \][/tex]
Factor out [tex]\(p^2\)[/tex]:
[tex]\[ 7m p^2 (p + 3m - 2p^2) \][/tex]
So, the factorised form is:
[tex]\[ \boxed{7mp^2(p + 3m - 2p^2)} \][/tex]
### d. Factorise [tex]\(7(y + 2) + x(2 + y)\)[/tex]
Observe that we can regroup and rearrange terms:
[tex]\[ 7(y + 2) + x(2 + y) \][/tex]
Reorder [tex]\(x(2 + y)\)[/tex] as [tex]\(x(y + 2)\)[/tex] (since addition is commutative):
[tex]\[ 7(y + 2) + x(y + 2) \][/tex]
Factor out the common term [tex]\((y + 2)\)[/tex]:
[tex]\[ (y + 2)(7 + x) \][/tex]
So, the factorised form is:
[tex]\[ \boxed{(y + 2)(x + 7)} \][/tex]
### e. Factorise [tex]\(7(h + k)^2 - 21(h + k)\)[/tex]
First, observe the common factor:
[tex]\[ 7(h + k)^2 - 21(h + k) \][/tex]
Factor out [tex]\(7(h + k)\)[/tex]:
[tex]\[ 7(h + k)[(h + k) - 3] \][/tex]
Simplify inside the brackets:
[tex]\[ 7(h + k)(h + k - 3) \][/tex]
So, the factorised form is:
[tex]\[ \boxed{7(h + k)(h + k - 3)} \][/tex]
These are the complete factorisations of the given polynomials.
### a. Factorise [tex]\(x^2 - 8xy\)[/tex]
First, we look for a common factor in both terms:
[tex]\[ x^2 - 8xy \][/tex]
Both terms have a common factor of [tex]\(x\)[/tex]:
[tex]\[ x(x - 8y) \][/tex]
Thus, the factorised form is:
[tex]\[ \boxed{x(x - 8y)} \][/tex]
### b. Factorise [tex]\(6q^3 - 2q + 2q^4\)[/tex]
First, we arrange the terms in standard form:
[tex]\[ 2q^4 + 6q^3 - 2q \][/tex]
Observe the common factor in all terms, which is [tex]\(2q\)[/tex]:
[tex]\[ 2q(q^3 + 3q^2 - 1) \][/tex]
So, the factorised form is:
[tex]\[ \boxed{2q(q^3 + 3q^2 - 1)} \][/tex]
### c. Factorise [tex]\(7mp^3 + 21m^2p^2 - 14mp^4\)[/tex]
First, identify the common factor:
[tex]\[ 7mp^3 + 21m^2p^2 - 14mp^4 \][/tex]
Factor out [tex]\(7m\)[/tex]:
[tex]\[ 7m(p^3 + 3m p^2 - 2p^4) \][/tex]
Factor out [tex]\(p^2\)[/tex]:
[tex]\[ 7m p^2 (p + 3m - 2p^2) \][/tex]
So, the factorised form is:
[tex]\[ \boxed{7mp^2(p + 3m - 2p^2)} \][/tex]
### d. Factorise [tex]\(7(y + 2) + x(2 + y)\)[/tex]
Observe that we can regroup and rearrange terms:
[tex]\[ 7(y + 2) + x(2 + y) \][/tex]
Reorder [tex]\(x(2 + y)\)[/tex] as [tex]\(x(y + 2)\)[/tex] (since addition is commutative):
[tex]\[ 7(y + 2) + x(y + 2) \][/tex]
Factor out the common term [tex]\((y + 2)\)[/tex]:
[tex]\[ (y + 2)(7 + x) \][/tex]
So, the factorised form is:
[tex]\[ \boxed{(y + 2)(x + 7)} \][/tex]
### e. Factorise [tex]\(7(h + k)^2 - 21(h + k)\)[/tex]
First, observe the common factor:
[tex]\[ 7(h + k)^2 - 21(h + k) \][/tex]
Factor out [tex]\(7(h + k)\)[/tex]:
[tex]\[ 7(h + k)[(h + k) - 3] \][/tex]
Simplify inside the brackets:
[tex]\[ 7(h + k)(h + k - 3) \][/tex]
So, the factorised form is:
[tex]\[ \boxed{7(h + k)(h + k - 3)} \][/tex]
These are the complete factorisations of the given polynomials.
