Algebra 2

To answer the first queston, first factor the polynomial function.

\(y={x}^{3}-{3x}^{2}+16x-48\)

\(y=({x}^{3}-{3x}^{2})+(16x-48)\)

\(y={x}^{2}(x-3)+(16x-48)\)

\(y={x}^{2}(x-3)+16(x-3)\)

\(y=({x}^{2}+16)\times(x-3)\)

\(y=({x}^{2}+{4}^{2})\times(x-3)\)

\(y=(x-4)\times(x+4)\times(x-3)\)

Now set each x value to zero and solve

\(x-4=0\)

\(x-4+4=0+4\)

\(x-0=0+4\)

\(x=0+4\)

\(x=4\)

\(x+4=0\)

\(x+4-4=0-4\)

\(x+0=0-4\)

\(x=0-4\)

\(x=-4\)

\(x-3=0\)

\(x-3+3=0+3\)

\(x-0=0+3\)

\(x=0+3\)

\(x=3\)

Now plug each answer to the original polynomial function to see if the answers will fit in the original polynomial function.

\(y={x}^{3}-{3x}^{2}+16x-48\)

\(0={x}^{3}-{3x}^{2}+16x-48\)

\(0={4}^{3}-3({4}^{2})+16(4)-48\)

\(0=64-3({4}^{2})+16(4)-48\)

\(0=64-3(16)+16(4)-48\)

\(0=64-48+16(4)-48\)

\(0=64-48+64-48\)

\(0=16+64-48\)

\(0=80-48\)

\(0≠32\)

\(y={x}^{3}-{3x}^{2}+16x-48\)

\(0={x}^{3}-{3x}^{2}+16x-48\)

\(0={(-4)}^{3}-{3(-4)}^{2}+16(-4)-48\)

\(0=-64-{3(-4)}^{2}+16(-4)-48\)

\(0=-64-3(16)+16(-4)-48\)

\(0=-64-48+16(-4)-48\)

\(0=-64-48+(-64)-48\)

\(0=-112+(-64)-48\)

\(0=-176-48\)

\(0≠-224\)

\(y={x}^{3}-{3x}^{2}+16x-48\)

\(0={x}^{3}-{3x}^{2}+16x-48\)

\(0={3}^{3}-{3(3}^{2})+16(3)-48\)

\(0=27-{3(3}^{2})+16(3)-48\)

\(0=27-3(9)+16(3)-48\)

\(0=27-27+16(3)-48\)

\(0=27-27+48-48\)

\(0=0+48-48\)

\(0=48-48\)

\(0=0\)

The only real answer is x=3.  Becasue x=4 and x=-4 are not real answers, that means they are imaginary answers.  This means that B is the correct answer.

I know how to solve the second question but, because I do not know how to type it out on this forum, I will leave that to someone who does know how to answer the question and how to type it out on this forum.

Part (1).  Not necessary to do all the above work. -tip ; always do a quick sketch to see that

When x is  negative and large , y is  negative and larger still.  When x is positive and large,y is positive and larger still.( The cubic term eventually outweighs all the others as x gets larger and larger.)

When x is  zero,  y =   -48.

so we know that  the function increases from negative infinity in the  3rd quadrant,crosses the y-axis at the point

( 0,-48)  and then keeps increasing. So it only has one real root.

A cubic has 3 roots,so the other two roots must be complex. (We can say more about these,but it's not asked for here.)

Part(2)

we can write a function as the product of some polynomial P,its quotient Q,and a remainder.

ie  y = P(x-2) + R   or f(x) = P(x-2) + R

so  4x^3-5x^2+3x-1  = P(x-2) + R

then f(2)  =R       = 17

The polynomial function y = x^3 -3x^2 + 16x - 48 has only one non-repeated x-intercept.

What do you know about the complex zeros of the function?

Degree 3
The zero set of discriminant of the cubic  \({\textstyle ax^{3}+bx^{2}+cx+d\,} \) has discriminant
  \( {\textstyle b^{2}c^{2}-4ac^{3}-4b^{3}d-27a^{2}d^{2}+18abcd\,.}\)
The discriminant is zero if and only if at least two roots are equal.
If the coefficients are real numbers, and the discriminant is not zero,
the discriminant is positive if the roots are three distinct real numbers,
and negative if there is one real root and two complex conjugate roots.

\(\begin{array}{|rcll|} \hline && b^{2}c^{2}-4ac^{3}-4b^{3}d-27a^{2}d^{2}+18abcd \quad |\quad a=1\quad b= -3 \quad c = 16\quad d = -48 \\\\ &=& (-3)^{2}\cdot(16)^{2}-4\cdot 1 \cdot(16)^{3}-4\cdot(-3)^{3}\cdot(-48)-27\cdot 1^{2}\cdot(-48)^{2}+18\cdot 1 \cdot (-3) \cdot (16) \cdot (-48) \\ &=& 9\cdot 16^{2}-4\cdot 16^{3}-4\cdot (-27) \cdot (-48) -27\cdot(-48)^{2}+18\cdot 3 \cdot (16) \cdot 48 \\ &=& 9\cdot 256-4\cdot 4096-4\cdot 1296 -27\cdot 2304 + 41472 \\ &=& 2304 - 16384 - 5184 -62208 + 41472 \\ &=& -40000 \\ \hline \end{array}\)

discriminant  = -40000 < 0  there is one real root and two complex conjugate roots.