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Agata [3.3K]
3 years ago
14

Have you done any factoring with variables? Which property (Commutative, Distributive, or Associative) do you think is most invo

lved in factoring? Do you think a polynomial can be a factor of another polynomial? Answer in full sentences.
Mathematics
2 answers:
vazorg [7]3 years ago
7 0
Idk how im supposed to answer this or whatevs bu
yes i have done factoring with variables.
distrubutive is the most involved in factoring method as i can tell.
i dont think ive studied this before or i mightve forgoten but im sure a polynomial can be factored into another one.
and yea. hope this helped in sum way
ASHA 777 [7]3 years ago
3 0

⇒x²y³=x*x*y*y*y=x*y*x*y*y=y*x*y*x*y--Writing factors in another way

⇒xy²z²=x*y*y*z*z=y*x*z*y*z=z*x*y*x*y*z---Writing factors in another way.

⇒xy+xy²=xy(1+y)

⇒y²z²-yzx=yz(yz-x)

Here ,we have used law of indices first, then used Commutativity , Associativity and Distributivity to factor Polynomial.

1.\rightarrow x^{a+b}=x^a \times x^b\\\\2.\rightarrow x \times y=y \times x----\text{Commutative Property}\\\\3.\rightarrow x \times (y \times z)= (x \times y) \times z=(x \times z) \times y---\text{Associaitive Property}

⇒Distributive Property of Multiplication with respect to addition and Subtraction

  →a×(b+c)=a×b+a×c

 → a×(b-c)=a×b-a×c

⇒Yes, A polynomial can be factor of another polynomial.

For Example:

⇒ x²-a², has two factors equal to , (x+a) and (x-a), where , x²-a², is a quadratic polynomial and two of it's factors are ,(x+a) and (x-a) which is also a polynomial of degree 1.

 

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Step-by-step explanation:

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3 years ago
On a coordinate plane, a parabola opens up. It goes through (negative 2, 4), has a vertex at (0.25, negative 6), and goes throug
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The statements true about the the function f(x) = 2x2 – x – 6 are-

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<h3>What is vertex of parabola?</h3>

The vertex of parabola is the point at the intersection of parabola and its line of symmetry.

Now the given function is,

f(x) = 2x^2 – x – 6

Also, it is given that the vertex is located at (0.25, -6)  and the parabola opens up, the function has two x-intercepts.

Comparing the given function with standard form,

f(x) = a x^2 bx + c

By comprison we get,

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Now, x-coordinate of vertex is given as,

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Put the value of x in given function, so y-coordinate of the vertex is given as,

f(1/4) = 2(1/4)² - 1/4 - 6

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Hence, The statements true about the the function f(x) = 2x2 – x – 6 are-

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  • The function has two x-intercepts.

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