Answer:
3x³ + x² + x + 1
= 3x³ - 3x² + 4x² - 4x + 5x - 5 + 6
= 3x²(x - 1) + 4x(x - 1) + 5(x - 1) + 6
= (x - 1)(3x² + 4x + 5) + 6
but 3x³ + x² + x + 1 = (x - 1).Q(x) + R
=> Q(x) = 3x² + 4x + 5
R = 6
Answer:
a) 
b) 
c) 
Step-by-step explanation:
<u>For the question a *</u> you need to find a polynomial of degree 3 with zeros in -3, 1 and 4.
This means that the polynomial P(x) must be zero when x = -3, x = 1 and x = 4.
Then write the polynomial in factored form.

Note that this polynomial has degree 3 and is zero at x = -3, x = 1 and x = 4.
<u>For question b, do the same procedure</u>.
Degree: 3
Zeros: -5/2, 4/5, 6.
The factors are

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<u>Finally for the question c we have</u>
Degree: 5
Zeros: -3, 1, 4, -1
Multiplicity 2 in -1

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30. Set up a proportion of 6/2 and x/10. cross multiply 6 times 10 and you get 60. divide 60 by 2=30
Answer:
The 4th graph
Step-by-step explanation:
To determine which graph corresponds to the f(x) = \sqrt{x} we will start with inserting some values for x and see what y values we will obtain and then compare it with graphs.
f(1) = \sqrt{1} = 1\\f(2) = \sqrt{2} \approx 1.41\\f(4) = \sqrt{4} = 2\\f(9) = \sqrt{9} = 3
So, we can see that the pairs (1, 1), (2, 1.41), (4, 2), (3, 9) correspond to the fourth graph.
Do not be confused with the third graph - you can see that on the third graph there are also negative y values, which cannot be the case with the f(x) =\sqrt{x}, the range of that function is [0, \infty>, so there are only positive y values for f(x) = \sqrt{x}