contestada

10. Use synthetic division to completely factor y = x3 - 4x2 - 9x + 36 by x + 3.
O.
y = (x + 3)(x - 4)(x - 3)
y = (x + 3)(x - 4)(x + 3)
y = (x + 3)(x + 4)(x - 3)
y = (x + 3)(x + 4)(x + 3)

Respuesta :

caylus

Answer: A

Step-by-step explanation:

[tex]P(x)= x^3 - 4x^2 - 9x + 36 \\D(x)=x + 3\\\\\begin{array}{|c|c|c|c|c|}&x^3&x^2&x&1\\---&---&---&---&---\\&1&-4&-9&36\\x=-3&&-3&21&-36\\---&---&---&---&---\\&1&-7&12&0\\\end{array}\\\\\\P(x)= x^3 - 4x^2 - 9x + 36 \\=(x+3)(x^2-7x+12)\\=(x+3)(x^2-3x-4x+12)\\=(x+3)(x(x-3)-4(x-3))\\=(x+3)(x-3)(x-4)\\\\Answer \ A[/tex]

r3t40

Start with,

[tex](x^3-4x^2-9x+36)\div(x+3)[/tex]

Look at first terms, [tex]x^3[/tex] and [tex]x[/tex] in divisor, with what do you have to multiply [tex]x[/tex] so that you end up with [tex]x^3[/tex]? The answer is [tex]x^2[/tex].

So you write,

[tex](x^3-4x^2-9x+36)\div(x+3)=x^2[/tex]

[tex]-(x^3+3x^2)[/tex] (because you are subtracting [tex]x^2(x+3)[/tex].

Which becomes when u subtract columns,

[tex](-7x^2-9x+36)\div(x+3)=x^2[/tex]

Now repeat the procedure,

[tex](-7x^2-9x+36)\div(x+3)=x^2-7x[/tex]

[tex]-(-7x^2-21x)[/tex]

which becomes

[tex](12x+36)\div(x+3)=x^2-7x[/tex]

And again,

[tex](12x+36)\div(x+3)=x^2-7x+12[/tex]

[tex]-(12x+36)[/tex]

becomes,

[tex]0=x^2-7x+12[/tex]

Since the RHS is 0 that means [tex](x+3)[/tex] is a factor.

Now you can factor [tex]x^2-7x+12=(x-4)(x-3)[/tex] because [tex]-4\cdot-3=12[/tex] and [tex]-4+(-3)=-7[/tex].

Combine the factors and get [tex](x+3)(x-4)(x-3)[/tex].

Hope this helps :)

Otras preguntas