briefly explain the process by which you would determine whether or not x-6 a factor of 3x^3-16x^2-72

2 answers:

5 0

Using the remainder theorem, substitute 6 in for any x in the equation and it will equal the remainder if it has been divided by (x-6). If the remainder is zero, then it would have divided evenly...making it a factor. If it equals anything but zero then it would not be a factor

storchak [24]3 years ago

3 0

X-6=0,x=6
put x=6 in the given eq. $3 x^{3} -16 x^{2} -72=3( 6^{3} )-16( 6^{2} )-72
=3(216)-16(36)-72
$
=648-576-72
=648-648
=0
hence x-6 is a factor of the given polynomial

You might be interested in

Help me with math, please

Alex777 [14]

Number 28 would be y= -52x+148

8 0

3 years ago

1. If a polynomial of Degree 4 has an imaginary zero at -2i and a real zero at 5, how many imaginary zeros does it have?

Vadim26 [7]

Answer:

Complex roots for polynomials equations will always come in conjugate pairs. Since 1-2i is a root, 1+2i will also be a root. (Just switch the sign of the imaginary part to its opposite.)

Step-by-step explanation:

So far we have 1-2i and 1+2i as roots. We need two more to make a 4th degree polynomial.

x=1 with multiplicity of 2 means that we count the root twice when we write down the polynomial: (x - 1)(x - 1) = (x - 1)2

Putting it all together, we can construct our polynomial of degree 4 as

y = f(x) = (x - 1)(x - 1)[x - (1-2i)][x - (1+2i)]

Multiply the factors (FOIL):

[x - (1-2i)][x - (1+2i)] =

x2 - (1+2i)x - (1-2i)x + (1-2i)(1+2i) =

x2 - x - 2ix - x + 2ix + (1 + 2i - 2i + 4) =

x2 - 2x + 5

(x - 1)(x - 1) = x2 - x - x + 1 = x2 - 2x + 1

Now multiply the two underlined expressions: