Question

Using identity ( x+ a )(x + b) = x^2 + ( a + b )x + ab. Find
( 3x + 5 ) ( 3x – 1 )

Answer 1

Answer:

(3x)²+(5-1)3+5.-1

9x²+12-5

9x²+7

hope this helps u

Step-by-step explanation:

Answer 2

$\large \bf \sf{Algebraic \: \: identities:-}$Algebraic identities are equalities that holds for any value of its variable. Algebraic identities are used for solving algebraic expressions and also for factorisation of polynomials.

$\large \sf \red{List \: of \: Indenties:- }$

$\sf{(i) \: (a + b {)}^{2} = {a}^{2} + {b}^{2} + 2(ab) }$

$\sf{(ii) \: ( a - b{)}^{2} = {a }^{2} - 2(ab) + {b}^{2} }$

$\sf{(iii) \: ( {a}^{2} - {b}^{2}) } = ( a+ b)( a- b)$

$\sf{(iv) \: (x + a)(x + b) = {x}^{2} + (a + b)x + ab }$

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

$\sf{(vi) \: ({a} + {b} {)}^{3} = {a}^{3} + {b}^{3} + 3ab (a + b) }$

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

$\sf{(viii) \: {a}^{3} + {b}^{3} + {c}^{3} - 3abc= (a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} -ab - bc - ca) }$

$\huge \underline \mathtt \red{Question:-}$

$\sf \color{navy}{Solve:- (3x+5)(3x-1)}$

$\huge \underline \mathtt \blue{Required \: Answer:-}$

$\sf{Using \: the \: identity:-} \\ \sf{( x+ a )(x + b) = x^2 + ( a + b )x + ab}$

So here, x= 3x , a= 5, b= 1

$\sf(3x + 5)(3x - 1)$

$\sf{ = >(3x {)}^{2} + (5 + 1)3x + (5 \times 1)}$

$\sf = > 9 {x}^{2} +( 6)3x + 5$

$\sf{ = > {9x}^{2} + 18x + 6 }$

$\large\sf \red{Hope \: it \: help s \: you.....}$