Answer:
Two complex roots.
Step-by-step explanation:
F(x)=2x^4 +5x^3 - x^2 +6x-1
is a polynomial in x of degree 4.
Hence F(x) has 4 roots. There can be 0 or 2 or 4 complex roots to this polynomial since complex roots occur in conjugate pairs.
Use remainder theorem to find the roots of the polynomial.
F(0) = -1 and F(1) = 2+5-1+6-1 = 11>0
There is a change of sign in F from 0 to 1
Thus there is a real root between 0 and 1.
Similarly by trial and error let us find other real root.
F(-3) = -1 and F(-4) = 94
SInce there is a change of sign, from -4 to -3 there exists a real root between -3 and -4.
Other two roots are complex roots since no other place F changes its sign
Answer:
Option 1 is correct.
Step-by-step explanation:
Given the equation 
we have to choose the best statement describes the above equation.
→ (1)
As, the highest degree of its monomials i.e individual terms with non-zero coefficients is 2.
⇒ Degree of above equation is 2.
hence, the given equation is quadratic equation.
The general form of quadratic equation is
In variable u: 
→ (2)
Now, compare equation (1) with (2), we say that
The equation is quadratic in form because it can be rewritten as a quadratic equation with u substitution u = (x + 5).
Option 1 is correct.
Hello! And thank you for your question!
First we are going to expand the equation:
<span>−2<span>m^<span><span>2</span><span></span></span></span>−2mn+8m−10m+10n+nm+4<span>n<span><span>^2</span><span></span></span></span>−5n
Then we are going to combine like terms:
</span><span><span>−2<span>m<span><span>^2</span><span></span></span></span>+(−2mn+mn)+(8m−10m)+(10n−5n)+4<span>n<span><span>^2</span><span></span></span></span></span>
</span>
Then finally, simplify:
−2<span>m<span><span>^2</span><span></span></span></span>−mn−2m+5n+4<span>n<span>^<span>2
Final Answer:
</span></span></span>
−2m^2−mn−2m+5n+4n^2
