Can some one solve questions 6&7 ? If u can only solve one, it is also fine! Thanks a lot !!!!!?

A polynomial f(x) of degree 4 has a quadratic factor of (x² - x + 3). One of the roots of f(x) = 0 is -3. Given that f(x) leaves a remainder of 60 and -36 when divided by (x - 1) and (x + 2) respectively, find:

the remaining factor of f(x):

We have one quadratic factor and one other root, which means it's also has a factor of (x + 3)

This leaves the final factor unknown, which I'll call (x - z) where z is the final root.

So we have:

(x² - x + 3)(x + 3)(x - z) = 0

This then is also multiplied by an unknown constant, which I'll call "a": When a is multiplied by 0 in the right side, we still have 0:

a(x² - x + 3)(x + 3)(x - z) = 0

Now the remainders when divided by (x - k) is the same answer if we set x = k and simplify.

So when divided by (x - 1) the remainder is 60, that means f(1) = 60.
same for f(-2) = -36

So knowing that, we can put this in our general form above to end up with two equations and two unknowns:

f(x) = a(x² - x + 3)(x + 3)(x - z)
60 = a(1² - 1 + 3)(1 + 3)(1 - z) and f(-2) = a((-2)² - (-2) + 3)(-2 + 3)(-2 - z)

Simplify both to get:

60 = a(1 - 1 + 3)(4)(1 - z) and -36 = a(4 + 2 + 3)(1)(-2 - z)
60 = a(3)(4)(1 - z) and -36 = a(9)(1)(-2 - z)
60 = 12a(1 - z) and -36 = 9a(-2 - z)

Dividing both sides by 12 and 9:

5 = a(1 - z) and -4 = a(-2 - z)

Solve for a in terms of z in one equation and subsitutute:

a = 5 / (1 - z)

-4 = a(-2 - z)
-4 = (5 / (1 - z))(-2 - z)
-4 = 5(-2 - z) / (1 - z)
-4(1 - z) = 5(-2 - z)
-4 + z = -10 - 5z
6z = -6
z = -1

Now that we have z, we can solve for a:

a = 5 / (1 - z)
a = 5 / (1 - (-1))
a = 5 / (1 + 1)
a = 5 / 2

So now the equation we have is:

f(x) = a(x² - x + 3)(x + 3)(x - z)
f(x) = (5/2)(x² - x + 3)(x + 3)(x + 1)

Now expand this to a polynomial:

f(x) = (5/2)(x² - x + 3)(x² + 4x + 3)
f(x) = (5/2)(x⁴ + 4x³ + 3x² - x³ - 4x² - 3x + 3x² + 12x + 9)
f(x) = (5/2)(x⁴ + 3x³ + 2x² + 9x + 9)

and finally:

f(x) = (5/2)x⁴ + (15/2)x³ + 5x² + (45/2)x + (45/2)

Part 2: the number of real foots: We have two real roots already. Just need to use the quadratic factor's discriminant to see if it maps to two real roots or two complex roots:

b² - 4ac
(-1)² - 4(1)(3)
1 - 12
-11

It's negative, so two complex roots. So two real roots in total.

Part 3: The remainder when f(x) is divided by x? In terms of (x - k) from above, k = 0, so when x = 0, what it simplifies to is the remainder. This leaves only the constant term of:

45/2

6 and 7 look hard and alot of work