Respuesta :
Answer:
Q1 - D. f(x) = [tex]x^4-9x^2-50x-150[/tex]
Q13 - A. ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24.
Q14 - A. [tex]7x+6+4x^{2}[/tex]
Step-by-step explanation:
Question 1:
We know that rational roots always occurs in pairs. So, the zeros of the function will be 5, -3, -1+3i, -1-3i
So, the factored form is [tex](x-5)(x+3)(x+1-3i)(x+1+3i)=0[/tex]
i.e. [tex](x^2-2x-15)(x+1-3i)(x+1+3i)=0[/tex]
i.e. [tex](x^3-x^2-3ix^2-17x+6ix-15+45i)(x+1+3i)=0[/tex]
i.e. [tex]x^4-9x^2-50x-150=0[/tex]
Hence, the polynomial function is [tex]f(x)=x^4-9x^2-50x-150[/tex].
Question 13:
Rational Zeros Theorem states that 'If p(x) is a polynomial with integer coefficients and if [tex]\frac{p}{q}[/tex] is a zero of p(x) = 0. Then, p is a factor of the constant term of p(x) and q is a factor of the leading coefficient of p(x)'.
Let, [tex]\frac{p}{q}[/tex] is a zero of [tex]x^3-7x^2+9x-24=0[/tex]. Then, p is a factor of -24 and q is a factor of 1.
Thus, possible values of p = ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24 and q = ±1
This gives, possible values of [tex]\frac{p}{q}[/tex] are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24.
Question 14:
We have, f(x) = 7x + 6 and g(x) = [tex]4x^{2}[/tex]
Then, (f+g)(x) = f(x) + g(x) = 7x + 6 + [tex]4x^{2}[/tex]
So, (f+g)(x) = [tex]7x+6+4x^{2}[/tex]
