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3 years ago

14

What is the solution of

1 answer:

marta [7]3 years ago

8 0

Answer:

$x$

Step-by-step explanation:

Firstly, we know that the solution cannot contain $\leq or \geq$ as when x=3, the equation is undefined. This cuts the options in half

When we plug in a value that is less than 3, we always get a value less than 1. This means that when added to 3, we will always get a value less than 4.

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P(x) = x^4+ax^3-x^2+bx-12 has factors of (x-2) and (x+1). Solve the equation P(x)=0

Feliz [49]

Answer:

because P(x) has factors of (x-2) and (x+1) => x = 2 and x = -1 are the solutions of P(x)

so we have:

$\left \{ {{2^{4}+2^{3}a-2^{2}+2b-12 =0} \atop {1-a-1-b-12=0}} \right.\\=>\left \{ {{8a+2b=0} \atop {a+b=-12}} \right.\\\left \{ {{4a+b=0} \atop {a+b=-12}} \right.\\\left \{ {{a=4} \atop {b=-16}} \right.$

=> $P(x)=x^{4} + 4x^{3}-x^{2} -16x-12$

With P(x) = 0

=> $x^{4}+4x^{3}-x^{2} -16x-12 = 0$

=> ........

Step-by-step explanation:

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3 years ago

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Answer:

Step-by-step explanation:

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