Question

May anyone please help. it will mean alot to me ​

Answer 1

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`1.\bf \quad(m + 11)(m - 11)`

Using the algebraic identity :

`\sf(a + b)(a - b) = {a}^(2) - {b}^(2)`

Therefore,

`\sf \longrightarrow \: {m}^(2) - {11}^(2)`

`\sf \longrightarrow \: {m}^(2) - 121`

i.e Product of sum and difference.

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`\bf 2. \quad(y + 12)(y - 5)`

`\sf \longrightarrow \: y(y - 5) + 12(y - 5)`

`\sf \longrightarrow \: {y}^(2) - 5y + 12y - 60`

`\sf \longrightarrow \: {y}^(2) + 7y - 60`

i.e, Product of two binomials.

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`\bf3. \quad \: (x + 15)(x + 15)`

`\sf \longrightarrow \: {(x + 15)}^(2)`

Using the algebraic identity,

`\sf {(a + b)}^(2) = {a}^(2) + {b}^(2) + 2ab`

`\sf \longrightarrow \: {x}^(2) + 225 + 30x`

i.e, Square of a binomial.

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`\bf \: 4. \quad \: {(x - 7)}^(3)`

Using the algebraic identity,

`\sf \:{(a - b)}^(3) = {a}^(3) - {b}^(3) - 3ab(a - b)`

`\sf \longrightarrow \: {x}^(3) - 343 - 21 {x}^(2) + 147x`

i.e, Cube of a binomial.

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`\bf \: 5. \quad \: {(6u + 7v + 8w)}^(2)`

Using the algebraic identity,

`\sf{(a + b + c)}^(2) = {a}^(2) + {b}^(2) + {c}^(2) + 2ab + 2bc + 2ca`

`\sf \longrightarrow \: 36 {u}^(2) + 49 {v}^(2) + 64 {w}^(2) + 84uv + 112vw + 96wu`

i.e, Square of a trinomial.

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