Question

factorise 9(p-q)^2 -4(2p+q)^2

Answer 1

Step-by-step explanation:

solution

Given

${9(p - q)}^{2} - \: 4 {(2p + q)}^{2} \\ \\ \\ open \: the \: bractest \: using \: formula \\ \\ 9( {p}^{2} - 2pq + {q}^{2} ) - 4(4 {p}^{2} + 4pq + {q}^{2} ) \\ \\ 9 {p }^{2} - 18pq + 9 {q}^{2} - 16 {p }^{2} - 16pq - 4{q}^{2} \\ \\ -7 {p }^{2} - 34pq + 5 {q}^{2} \\ \\ 7 {p}^{2} + 34pq - 5 {q}^{2} \\ \\ factorise \: this \: we \: get \\ \\ 7 {p }^{2} + 35pq - pq - 5 {q}^{2} \\ \\ take \: a \: commen \\ \\ 7p(p + 5q) - q(p + 5q) \\ \\ (p + 5q) \: (7p - q)$

Answer 2

$9(p-q)^{2} -4(2p+q)^{2}=(7p-q)(-p-5q)$.

Step-by-step explanation:

We have,

$9(p-q)^{2} -4(2p+q)^{2}$

=$(3(p-q))^{2} -(2(2p+q))^{2}$

$=[3(p-q)+2(2p+q)][[3(p-q)-2(2p+q)][$

[Since,$a^{2} -b^{2} =(a+b)(a-b)$]

$=(3p-3q+4p+2q)(3p-3q-4p-2q)$

$=(7p-q)(-p-5q)$

The value of $9(p-q)^{2} -4(2p+q)^{2}$ is equal to $(7p-q)(-p-5q).$

Hence, $9(p-q)^{2} -4(2p+q)^{2}=(7p-q)(-p-5q)$.