A polynomial of degree n with roots
can be represented as:
, where A is a constant (The same "a" that appears outside the parentheses in the field where you will type the polynomial). In this question, the number of roots is 4, because the degree of f(x) is 4. If a polynomial has only real coefficients, the complex roots appear in pairs of conjugates. So, since
is a zero,
is a zero too. Then, we have the 4 zeros. Hence,
will be in the form:
Answer:
Step-by-step explanation:
It would be point B. Since y-intercept is when the x is at 0, the answer would be point B.
the quotient will be x+3 and remainder is x^2+10x-100
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Answer:
1.
centre(h,k)=(-13,9)
radius (r)=6
we have
equation of the circle is
(x-h)²+(y-k)²=r²
(x+13)²+(y-9)²=6²
x²+26x+169+y²-18y+81=36
x²+y²+26x-18y+169+81-36=0
x²+y²+26x-18y +214=0
is a required equation of the circle.
2.
centre(h,k)=(1,-1)
radius (r)=11
we have
equation of the circle is
(x-h)²+(y-k)²=r²
(x-1)²+(y+1)²=11²
x²-2x+1+y²+2y+1=121
x²+y²-2x+2y=121-2
x²+y²-2x+2y=119
is a required equation of the circle.