Suppose $p(x)$ is a monic cubic polynomial with real coefficients such that $p(3-2i)=0$ and $p(0)=-52$.
Determine $p(x)$ (in expanded form).
ah, actually I misspoke.
Since 3+2i is also a root, we have
p(x) = (x-(3-2i))(x-(3+2i))(x-a)
= (x^2-6x+13)(x-a)
So, now just solve for a:
p(0) = -13a = -52
a = 4
p(x) = (x^2-6x+13)(x-4) = x^3-10x^2+37x-52
Thanks so much :)
i would like to 1v1 someone in valorant drop your name and tagline
Chunky Kong
yes
I love how this started with math and out of nowhere came a 1v1 request
ur mom
To determine the polynomial $p(x)$, we need to use the given information about its roots and the fact that it has real coefficients.
Firstly, we are given that $p(3-2i) = 0$. This means that $3-2i$ is a root of the polynomial. Since $p(x)$ has real coefficients, its complex conjugate, $3+2i$, must also be a root.
So we have two roots: $x = 3-2i$ and $x = 3+2i$.
Now, let's use the fact that $p(x)$ is a cubic polynomial to write it in factored form:
$p(x) = (x - r_1)(x - r_2)(x - r_3)$
where $r_1$, $r_2$, and $r_3$ are the roots of $p(x)$. In this case, $r_1 = 3-2i$, $r_2 = 3+2i$, and $r_3$ is the other root we need to find.
Since $p(x)$ is monic (the leading coefficient is 1), we can also write it in expanded form as:
$p(x) = (x - r_1)(x - r_2)(x - r_3) = (x - 3 + 2i)(x - 3 - 2i)(x - r_3)$
Now, we can use the other given information that $p(0) = -52$. Since $p(0) = (0 - 3 + 2i)(0 - 3 - 2i)(0 - r_3)$, we can solve for $r_3$:
$-52 = (0 - 3 + 2i)(0 - 3 - 2i)(0 - r_3)$
Next, let's simplify the expression inside the parentheses:
$(0 - 3 + 2i)(0 - 3 - 2i) = (2i)(-2i) = 4$
So now we have:
$-52 = 4(0 - r_3)$
$-52 = -4r_3$
Dividing by -4, we get:
$13 = r_3$
Therefore, the polynomial $p(x)$ in expanded form is:
$p(x) = (x - 3 + 2i)(x - 3 - 2i)(x - 13)$
peter is 30 lbs lighter than john. So,
j + j-30 = 235
Odd; I expected an integer answer...
as for the polynomial, who knows? I see no question there. And LaTex doesn't do well here. And p(x) cannot be monic with real coefficients and have a complex root.