Answer:
false
Step-by-step explanation:
if is 6= you need to time by it self like: 6×6=is ur answer
There are many polynomials that fit the bill,
f(x)=a(x-r1)(x-r2)(x-r3)(x-r4) where a is any real number not equal to zero.
A simple one is when a=1.
where r1,r2,r3,r4 are the roots of the 4th degree polynomial.
Also note that for a polynomial with *real* coefficients, complex roots *always* come in conjugages, i.e. in the form a±bi [±=+/-]
So a polynomial would be:
f(x)=(x-(-4-5i))(x-(-4+5i))(x--2)(x--2)
or, simplifying
f(x)=(x+4+5i)(x+4-5i)(x+2)^2
=x^4+12x^3+77x^2+196x+164 [if you decide to expand]
61 is the composite Number your welcome
<h3>
Answer: x(x+1)(5x+9) </h3>
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Work Shown:
5x^3 + 14x^2 + 9x
x( 5x^2 + 14x + 9 )
To factor 5x^2 + 14x + 9, we could use the AC method and guess and check our way to getting the correct result.
A better way in my opinion is to solve 5x^2 + 14x + 9 = 0 through the quadratic formula

Then use those two solutions to find the factorization
x = -1 or x = -9/5
x+1 = 0 or 5x = -9
x+1 = 0 or 5x+9 = 0
(x+1)(5x+9) = 0
So we have shown that 5x^2 + 14x + 9 factors to (x+1)(5x+9)
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Overall,
5x^3 + 14x^2 + 9x
factors to
x(x+1)(5x+9)