Respuesta :

sam4040

Answer:

[tex](4p+3q)(4p+7q)=x^2+40pq+21q^2[/tex]

Step-by-step explanation:

[tex]Given\ (x+a)(x+b)=x^2+(a+b)x+ab\ ...............................(1)\\\\We\ have\ to \find\ (4p+3q)(4p+7q)\\\\Compare\ (4p+3q)(4p+7q)\ with\ (x+a)(x+b)\\\\x=4p\\\\a=3q\\\\b=7q\\\\Substitute\ these\ values\ in\ eq(1)\\\\(4p+3q)(4p+7q)=x^2+(3q+7q)\times 4p+3q\times 7q\\\\(4p+3q)(4p+7q)=x^2+10q\times 4p+21q^2\\\\(4p+3q)(4p+7q)=x^2+40pq+21q^2[/tex]

Otras preguntas