Answer:
Question 1
Given expression:
[tex]-2x^3 (4x^3+5x^6)[/tex]
Apply the distributive law [tex]a(b+c)=ab+ac[/tex]:
[tex]\implies -2x^3 \cdot4x^3+-2x^3 \cdot 5x^6[/tex]
Multiply the numbers:
[tex]\implies -8x^3 x^3-10x^3 x^6[/tex]
Apply exponent rule [tex]a^b \cdot a^c=a^{b+c}[/tex]
[tex]\implies -8x^6-10x^9[/tex]
In standard form we write it in order from the greatest degree:
[tex]\implies -10x^9-8x^6[/tex]
Therefore, as the polynomial is degree 9, it is a nonic polynomial.
End behaviors
As the leading coefficient is negative and the leading degree is odd:
[tex]f(x) \rightarrow + \infty \textsf{, as } x \rightarrow - \infty[/tex]
[tex]f(x) \rightarrow - \infty \textsf{, as } x \rightarrow + \infty[/tex]
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Question 2
Factor the given function to find the zeros
Given function:
[tex]f(x) =x^3+15x^2 +63x+49[/tex]
[tex]f(-1) =(-1)^3+15(-1)^2 +63(-1)+49=0[/tex]
Therefore [tex](x+1)[/tex] is a factor of the function
[tex]\implies x^3+15x^2 +63x+49=(x+1)(x^2+bx+49)[/tex]
[tex]=x^3+(b+1)x^2+(b+49)x+49[/tex]
Comparing coefficients ⇒ b = 14
[tex]\implies x^3+15x^2 +63x+49=(x+1)(x^2+14x+49)[/tex]
Factoring [tex](x^2+14x+49)[/tex]:
[tex]\implies (x^2+14x+49)=(x+7)(x+7)=(x+7)^2[/tex]
Therefore,
[tex]f(x)=(x+1)(x+7)^2[/tex]
To find the zeros, equate to the function zero:
[tex](x+1)(x+7)^2=0[/tex]
Therefore,
[tex]x+1=0 \implies x=-1 \textsf{ with multiplicity 1}[/tex]
[tex]x+7=0 \implies x=-7\textsf{ with multiplicity 2}[/tex]
