Question

PLEASE HELP I'M BEGGING YOU SHOW STEPS IF YOU CAN

Answer 1

#1

You need to simplify according to distribute law

• a(b+c)=ab+bc

$\\ \rm\longmapsto -2x^3(4x^3-5x^6)$

$\\ \rm\longmapsto -8x^{3+3}-10x^{6+3}$

$\\ \rm\longmapsto -8x^6-10x^9$

• Highest degree=9

#2

• We need the graph (attached)

Zeros are the x intercepts

• they are -7 and -1

Answer 2

Answer:

Question 1

Given expression:

$-2x^3 (4x^3+5x^6)$

Apply the distributive law $a(b+c)=ab+ac$:

$\implies -2x^3 \cdot4x^3+-2x^3 \cdot 5x^6$

Multiply the numbers:

$\implies -8x^3 x^3-10x^3 x^6$

Apply exponent rule  $a^b \cdot a^c=a^{b+c}$

$\implies -8x^6-10x^9$

In standard form we write it in order from the greatest degree:

$\implies -10x^9-8x^6$

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:

$f(x) \rightarrow + \infty \textsf{, as } x \rightarrow - \infty$

$f(x) \rightarrow - \infty \textsf{, as } x \rightarrow + \infty$

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Question 2

Factor the given function to find the zeros

Given function:

$f(x) =x^3+15x^2 +63x+49$

$f(-1) =(-1)^3+15(-1)^2 +63(-1)+49=0$

Therefore $(x+1)$ is a factor of the function

$\implies x^3+15x^2 +63x+49=(x+1)(x^2+bx+49)$

                                         $=x^3+(b+1)x^2+(b+49)x+49$

Comparing coefficients ⇒ b = 14

$\implies x^3+15x^2 +63x+49=(x+1)(x^2+14x+49)$

Factoring $(x^2+14x+49)$:

$\implies (x^2+14x+49)=(x+7)(x+7)=(x+7)^2$

Therefore,

$f(x)=(x+1)(x+7)^2$

To find the zeros, equate to the function zero:

$(x+1)(x+7)^2=0$

Therefore,

$x+1=0 \implies x=-1 \textsf{ with multiplicity 1}$

$x+7=0 \implies x=-7\textsf{ with multiplicity 2}$