Question

ayuda!!! Es para una hora​

Answer 1

Respuesta:

a) (6x+3)(7x-2)=

(6x+3)(7x−2)

=(6x+3)(7x+−2)

=(6x)(7x)+(6x)(−2)+(3)(7x)+(3)(−2)

$= 42 {x}^{2} - 12x + 21x - 6$

$= 42 {x}^{2} + 9x - 6$

b)

$(2 {x}^{2} + 1)(3x - 4)$

$= (2 {x}^{2} +1)(3x+−4)$

$= (2 {x}^{2} )(3x) + (2 {x}^{2})( - 4) + (1)(3x) + (1)( - 4)$

$=6 {x}^{3} −8 {x}^{2} +3x−4$

c)

$(7x + y)(2x + 5y)$

=(7x+y)(2x+5y)

=(7x)(2x)+(7x)(5y)+(y)(2x)+(y)(5y)

$= 14 {x}^{2} + 35xy + 2xy + 5 {y}^{2}$

$=14 {x}^{2} +37xy+5 {y}^{2}$

d)

$(3 {x}^{3} + 2 {x}^{2} - 3)( \times - 2) =$

$=(3 {x}^{2} )(x)+(3 {x}^{2} )(−2)+(2 {x}^{2} )(x)+(2 {x}^{2} )(−2)+(−3)(x)+(−3)(−2)$

$=3 {x}^{4} −6 {x}^{3} +2 {x}^{2} −4 {x}^{2} −3x+6$

$=3 {x}^{4} −4 {x}^{3} −4 {x}^{2} −3x+6$

e)

$( {x}^{2} {y}^{2} + 2 {x}^{2} - 3x {y}^{2} (xy + y + 2)$

$= {x}^{2} {y}^{2} +2 {x}^{2} +−{4}^{3} y+−4 {x}^{2} y+−8 {x}^{2}$

$=(−4 {x}^{3} y)+( {x}^{2} {y}^{2} )+(−4 {x}^{2} y)+(2 {x}^{2} +−8 {x}^{2} )$

$=−4 {x}^{3} y+ {x}^{2} {y}^{2} +−4 {x}^{2} y+−6 {x}^{2}$

f)

$(2 {x}^{2} - 3 {x}^{2} )( {x}^{2} - 1) =$

$=(2 {x}^{2} )( {x}^{2} )+(2 {x}^{2} )(−1)+(−9)( {x}^{2} )+(−9)(−1)$

$=2 {x}^{4} −2 {x}^{2} −9 {x}^{2} +9$

$=2 {x}^{4} −11 {x}^{2} +9$

g)

$(2 {x}^{3} - 3 {x}^{2} {y}^{4} + xy)(5x - 2) =$

$=(2 {x}^{3} )(5x)+(2 {x}^{3} )(−2)+(−3 {x}^{2} {y}^{4} )(5x)+(−3 {x}^{2} {y}^{4} )(−2)+(xy)(5x)+(xy)(−2)$

$=10{x}^{4} −4 {x}^{3} −15 {x}^{3} {y}^{4} +6 {x}^{2} {y}^{4} +5 {x}^{2} y−2xy$

$=−15 {x}^{3} {y}^{4} +6 {x}^{2} {y}^{4} +10 {x}^{4} −4 {x}^{3} +5 {x}^{2} y−2xy$

h)

$(4 {x}^{2} + 3x + 2)( {x}^{3} - 2{x}^{2} 1) =$

$=(4 {x}^{2} )(33)+(4 {x}^{2} )(−2 {x}^{21} )+(3x)(33)+(3x)(−2 {x}^{21} )+(2)(33)+(2)(−2 {x}^{21} )$

$=108 {x}^{2} −8 {x}^{23} +81x−6 {x}^{22} +54−4 {x}^{21}$

$=−8 {x}^{23} −6 {x}^{22} −4 {x}^{21} +108 {x}^{2} +81x+54$