Question

In the special Product, (x+7)(x-8)(x+6), the co-efficient of x^2 is --------------​

Answer 1

Given :

$\tt \large (x + 7)(x + 8)(x + 6)$

To Find :

The coefficient of x²

Before the Question let's know about the coefficients.

Coefficient is an integer that is multiplied with a single term or with a polynomial. (with sign).

Hope you catch it So, lets get started.

To find the coefficient of x² first we will multiply$\tt \large (x + 7)$with $\tt \large (x + 8)$

and then we will multiply the resultant with $\tt \large (x + 6)$

So,

$\tt \large (x + 7) \times (x + 8)$

$\tt \: ↳(x \times x + x \times 8 + 7 \times x + 7 \times 8)(x + 6)$

$\tt \: ↳( {x}^{2} + 8x + 7x + 56)(x + 6)$

$\tt \: ↳( {x}^{2} + 15x + 56)(x + 6)$

Now multiplying$\tt ( {x}^{2} + 15x + 56)$with $\tt \: (x + 6)$

So,

$↳ \tt \small ( {x}^{2} + 15x + 56) \times (x + 6)$

$↳ \tt \small ( {x}^{2} \times x + 15x \times x + 56 \times x + {x}^{2} \times 6 + 15x \times 6 + 56 \times 6 )$

$↳ \tt \small( {x}^{3} + 15 {x}^{2} + 56x + 6 {x}^{2} + 90x + 336)$

Proceeding with simple calculation

$↳ \tt \small ( {x}^{3} + 15 {x}^{2} + 6 {x}^{2} + 90x + 56x + 336)$

$↳ \tt \small ( {x}^{3} + 21 {x}^{2} + 146x + 336)$

Removing the bracket

$↳ \tt \small {x}^{3} + 21 {x}^{2} + 146x + 336$

Henceforth,

Clearly the coefficient of x² is $\sf \color{teal} + 21$

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