Question

evaluate using identities (4x+5)(4x-1)​

Answer 1

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✪QUESTION✪:-

evaluate using identities

(4x+5)(4x-1)

✪ANSWER✪:-

IDENTITY USED :

(x + a)(x + b) = x² + (a + b) x + ab

here ,

• x = 4x
• a = 5
• b = -1

➠(4x+5)(4x-1)

➠(4x)² +(5-1)(4x) + (5)(-1)

➠16x² + (4)(4x) -5

➠16x² + 16x -5

◆ ━━━━━━━━❪✪❫━━━━━━━━━ ◆

ALGEBRAIC IDENTITIES :

IDENTITY I:

(a + b)² = a² + 2ab + b²

IDENTITY II:

(a – b)² = a² – 2ab + b²

IDENTITY III:

a² – b²= (a + b)(a – b)

IDENTITY IV:

(x + a)(x + b) = x² + (a + b) x + ab

IDENTITY V:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

IDENTITY VI:

(a + b)³ = a³ + b³ + 3ab (a + b)

IDENTITY VII:

(a – b)³ = a³ – b³ – 3ab (a – b)

IDENTITY VIII:

a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca)

Answer 2

$\huge \orange{ \overbrace{ \red{\underbrace \color{blue} {\underbrace \color{black}{\colorbox{lime}{ {\red \: {☘Answer ☘ }}}}}}}}$

✪QUESTION✪:-

evaluate using identities

(4x+5)(4x-1)

✪ANSWER✪:-

IDENTITY USED :

(x + a)(x + b) = x² + (a + b) x + ab

here ,

x = 4x

a = 5

b = -1

➠(4x+5)(4x-1)

➠(4x)² +(5-1)(4x) + (5)(-1)

➠16x² + (4)(4x) -5

➠16x² + 16x -5

◆ ━━━━━━━━❪✪❫━━━━━━━━━ ◆

ALGEBRAIC IDENTITIES :

IDENTITY I:

(a + b)² = a² + 2ab + b²

IDENTITY II:

(a – b)² = a² – 2ab + b²

IDENTITY III:

a² – b²= (a + b)(a – b)

IDENTITY IV:

(x + a)(x + b) = x² + (a + b) x + ab

IDENTITY V:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

IDENTITY VI:

(a + b)³ = a³ + b³ + 3ab (a + b)

IDENTITY VII:

(a – b)³ = a³ – b³ – 3ab (a – b)

IDENTITY VIII:

a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca)