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3 years ago

What is the complete factorization of the polynomial below? x^3+ 5x^2- x-5

Mathematics

1 answer:

koban [17]3 years ago

7 0

Answer:

(x-1)(x+1)(x+5)

Step-by-step explanation:

x^3+5x^2 - x-5=

x^3-x+5x^2-5=

x(x^2-1) +5(x^2 - 1)=

(x^2- 1)(x+5)=(*)

(x-1)(x+1)(x+5)

(*) A^2 - B^2 =(A-B)(A +B)

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What two numbers can add to 4 but multiple to -12

olasank [31]

Answer:

<em>Numbers: 6 and -2</em>

Step-by-step explanation:

<u>Equations</u>

This question can be solved by inspection. It's just a matter of factoring 12 into two factors that sum 4. Both numbers must be of different signs and they are 6 and -2. Their sum is indeed 6-2=4 and their product is 6*(-2)=-12.

However, we'll solve it by the use of equations. Let's call x and y to the numbers. They must comply:

$x+y=4\qquad\qquad [1]$

$x.y=-12\qquad\qquad [2]$

Solving [1] for y:

$y=4-x$

Substituting in [2]

$x(4-x)=-12$

Operating:

$4x-x^2=-12$

Rearranging:

$x^2-4x-12=0$

Solving with the quadratic formula:

$\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

With a=1, b=-4, c=-12:

$\displaystyle x=\frac{-(-4)\pm \sqrt{(-4)^2-4(1)(-12)}}{2(1)}$

$\displaystyle x=\frac{4\pm \sqrt{16+48}}{2}$

$\displaystyle x=\frac{4\pm 8}{2}$

The solutions are:

$\displaystyle x=\frac{4+ 8}{2}=6$

$\displaystyle x=\frac{4- 8}{2}=-2$

This confirms the preliminary results.

Numbers: 6 and -2

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Step-by-step explanation:

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Step-by-step explanation:

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