Question

(b + 4)² + 4(b + 4) - 21​

Answer 1

Step-by-step explanation:

${(b + 4)}^{2} + 4(b + 4) - 21 \\ (b + 4)(b + 4) + 4(b + 4) - 21 \\ b(b + 4) + 4(b + 4) + 4(b + 4) - 21 \\ b(b + 4) + 8(b + 4) - 21 \\ {b}^{2} + 4b + 8b + 32 - 21 \\ {b}^{2} + 4b + 8b + 11 \\ {b}^{2} + 12b + 11 \\ {b}^{2} + 11b + b + 11 \\ b(b +1 1) + b + 11 \\ (b + 1)(b + 11)$

Thanx

Answer 2

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: {(b + 4)}^{2} + 4(b + 4) - 21$

Let assume that

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{ \: \: \boxed{ \rm{ \:b + 4 = x \: }}}$

So, above expression can be rewritten as

$\rm \: = \: {x}^{2} + 4x - 21$

Now, on splitting the middle terms, we get

$\rm \: = \: {x}^{2} + 7x - 3x - 21$

$\rm \: = \: x(x + 7) - 3(x + 7)$

$\rm \: = \: (x + 7)(x - 3)$

On substituting the value of x = b + 4, we get

$\rm \: = \: (b + 4 + 7)(b + 4 - 3)$

$\rm \: = \: (b + 11)(b + 1)$

Hence,

$\red{\boxed{ \rm{{(b + 4)}^{2} + 4(b + 4) - 21 = \: (b + 11)(b + 1)}}}$

$\rule{190pt}{2pt}$

Basic Concept :-

Splitting of middle terms :-

In order to factorize  ax² + bx + c we have to find numbers m and n such that m + n = b and mn = ac.

After finding m and n, we split the middle term i.e bx in the quadratic equation as mx + nx and get the required factors by grouping the terms.

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$