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99 points!!!
Factor each expression completely.
1. x^2-5x-24
2. 13x^2-52
3. Find the equation of a line through (-6, 3) and parallel to 3x-9y=14
4. Solve (5x+4)(3x-7)=0; {rational numbers}

Respuesta :

apologiabiology
try to factor the greatest common factor out of each then if you have
1x²+bx+c, then find what 2 numbers (let's call them r and w) mutliply to get c and add to get b
it will factor to (x+w)(x+r)

and differnce of 2 perfect squares: a²-b²=(a-b)(a+b)

and paralell lines have same slope

and if xy=0 then assume x and/or y=0




so

1.
what 2 numbers multiply to get -24 and add to get -5
-8 and 3
(x-8)(x+3)


2.
factor out 13 from each
13(x²-4)
differnce of 2 perfect squares
13(x²-2²)=13(x-2)(x+2)


3.
a nice hack is for ax+by=c, a paralell line will be ax+by=z where the values of a and b are the same for both equations

so 3x-9y=14
a paralell line will be
3x-9y=c
find c
subsitute the point (-6,3)
x=-6 and y=3
subsitute
3(-6)-9(3)=c
-18-27=c
-45=c
the equation is 3x-9y=-45


4.
for so since those  2 things multiply to get 0, set them both equal to 0 and solve

5x+4=0
minus 4 both sides
5x=-4
divide by 5 both sides
x=-4/5

3x-7=0
add 7 both sides
3x=7
divide by 3 both sides
x=7/3

solutions are x=-4/5 and 7/3







1. (x-8)(x+3)
2. 13(x-2)(x+2)
3. 3x-9y=-45 (or simplified would be x-3y=-15)
4. x=-4/5 and 7/3

#1

[tex] \sf \: {x}^{2} - 5x - 24 \\ \sf \: {x}^{2} + 3x - 8x - 24 \\ \sf \: x \times (x + 3) - 8(x + 3) \\ \sf \: (x - 8) \times (x + 3)[/tex]

➪Therefore the factors are: (x-8) & (x+3)

[tex]\red{ \rule{35pt}{2pt}} \orange{ \rule{35pt}{2pt}} \color{yellow}{ \rule{35pt} {2pt}} \green{ \rule{35pt} {2pt}} \blue{ \rule{35pt} {2pt}} \purple{ \rule{35pt} {2pt}}[/tex]

#2

[tex] \tt \: {13x}^{2} - 52 \\ \tt13( {x}^{2} - 4) \\ \tt13(x - 2) \times (x + 2)[/tex]

➪ Therefore the factors are: 13(x-2) & (x+2)

[tex]\red{ \rule{35pt}{2pt}} \orange{ \rule{35pt}{2pt}} \color{yellow}{ \rule{35pt} {2pt}} \green{ \rule{35pt} {2pt}} \blue{ \rule{35pt} {2pt}} \purple{ \rule{35pt} {2pt}}[/tex]

#3

  • To find this we will use the standard form of equation i.e. ax+by+c=0
  • We will substitute the given values of x and y and solve the equation for c

  • x = -6
  • y = 3

Substitute the values into 3x-9y=14

[tex] \bf 3( - 6) - 9(3) = c[/tex]

[tex] \bf \: - 3 \times 6 - 9 \times 3 = c[/tex]

[tex] \bf \: 3( - 6 - 9) = c[/tex]

[tex] \bf \: 3( - 15) = c[/tex]

[tex] \bf \: - 45 = c[/tex]

[tex] \boxed{ \bf \: c = - 45}[/tex]

➪ So, the equation parallel to 3x-9y=14 is: 3x-9y=-45

[tex]\red{ \rule{35pt}{2pt}} \orange{ \rule{35pt}{2pt}} \color{yellow}{ \rule{35pt} {2pt}} \green{ \rule{35pt} {2pt}} \blue{ \rule{35pt} {2pt}} \purple{ \rule{35pt} {2pt}}[/tex]

[tex] \rm \: (5x + 4)(3x - 7) = 0[/tex]

[tex] \rm \: 5x + 4 = 0 \\ \rm \: 3x - 7 = 0[/tex]

[tex] \rm \: 5x = - 4 \\ \rm 3x = 7[/tex]

[tex] \rm \: x_{1} = - \frac{4}{5} , x_{2} = \frac{7}{3} [/tex]

➪ The values of x1 and x2 are: -4/5 & 7/3

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