Math Snap

STEP 1

1. We are given the polynomial $p(x) = ax^3 + bx^2 + cx + 60$.
2. When $p(x)$ is divided by $x-1$, the remainder is 30.
3. When $p(x)$ is divided by $(x-3)(x+7)$, the quotient is $x-3$.
4. We need to find the remainder when $p(x)$ is divided by $(x-3)(x+7)$.
5. We need to determine how many real roots the equation $p(x) = 0$ has.

STEP 2

1. Use the Remainder Theorem to find conditions on $a, b, c$.
2. Use the division information to set up an equation for $p(x)$.
3. Solve for the remainder when $p(x)$ is divided by $(x-3)(x+7)$.
4. Analyze the polynomial to determine the number of real roots.

STEP 3

Use the Remainder Theorem for $x-1$:
$$p(1) = a(1)^3 + b(1)^2 + c(1) + 60 = 30$$ Simplify:
$$a + b + c + 60 = 30$$ $$a + b + c = -30$$

STEP 4

Given that the quotient when dividing by $(x-3)(x+7)$ is $x-3$, we can express $p(x)$ as:
$$p(x) = ((x-3)(x+7))(x-3) + R(x)$$ where $R(x)$ is the remainder, a linear polynomial since the divisor is quadratic:
$$R(x) = dx + e$$

STEP 5

Substitute $x = 3$ and $x = -7$ into $p(x)$ using the division form to find $d$ and $e$.
For $x = 3$:
$$p(3) = ((3-3)(3+7))(3-3) + d(3) + e = d(3) + e$$ For $x = -7$:
$$p(-7) = ((-7-3)(-7+7))(-7-3) + d(-7) + e = d(-7) + e$$ These give us two equations:
1. $3d + e = p(3)$
2. $-7d + e = p(-7)$

STEP 6

Use $p(x) = ax^3 + bx^2 + cx + 60$ to find $p(3)$ and $p(-7)$:
$$p(3) = a(3)^3 + b(3)^2 + c(3) + 60$$ $$p(-7) = a(-7)^3 + b(-7)^2 + c(-7) + 60$$ Solve the equations for $d$ and $e$.

STEP 7

Substitute the known values into the equations:
$$3d + e = 27a + 9b + 3c + 60$$ $$-7d + e = -343a + 49b - 7c + 60$$ Solve these simultaneous equations to find $d$ and $e$.

Analyze the polynomial $p(x) = ax^3 + bx^2 + cx + 60$ for real roots.
1. The degree of $p(x)$ is 3, so it can have at most 3 real roots.
2. Use the information from the remainder and division to check for possible roots.
The remainder when $p(x)$ is divided by $(x-3)(x+7)$ is:
$$R(x) = dx + e$$ The number of real roots of $p(x) = 0$ depends on the discriminant and the nature of the roots found.