Math / Algebra

Question39. Let $p(x)=a x^{3}+b x^{2}+c x+60$ . When $p(x)$ is divided by $x-1$ , the remainder is 30 . When $p(x)$ is divided by $(x-3)(x+7)$ , the quotient is $x-3$ . (a) Find the remainder when $p(x)$ is divided by $(x-3)(x+7)$ . xplain (b) How many real roots does the equation $p(x)=0$ have? Explain your answer. $\Rightarrow$ Example 14

Studdy Solution

STEP 1

1. We have 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 the coefficients.
2. Use the given quotient information to find the remainder when dividing by $(x-3)(x+7)$ .
3. Analyze the polynomial to determine the number of real roots.

STEP 3

Using the Remainder Theorem, substitute $x = 1$ into $p(x)$ to find the remainder:
$p(1) = a(1)^3 + b(1)^2 + c(1) + 60 = 30$
Simplify:
$a + b + c + 60 = 30$ $a + b + c = -30$

STEP 4

Given that the quotient when $p(x)$ is divided 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:
$R(x) = mx + n$

STEP 5

Substitute $x = 3$ and $x = -7$ into $p(x)$ to find $m$ and $n$ :
For $x = 3$ :
$p(3) = (3-3)(3+7)(3-3) + R(3) = R(3)$ $R(3) = m(3) + n = 3m + n$
For $x = -7$ :
$p(-7) = (-7-3)(-7+7)(-7-3) + R(-7) = R(-7)$ $R(-7) = m(-7) + n = -7m + n$

STEP 6

Since $R(x)$ is the remainder when divided by $(x-3)(x+7)$ , and the quotient is $x-3$ , we know:
$p(x) = (x-3)(x+7)(x-3) + mx + n$
Given the conditions, solve for $m$ and $n$ using the system of equations from Step 3.

STEP 7

Use the fact that $p(x)$ has the remainder 30 when divided by $x-1$ to find the values of $m$ and $n$ .
$p(1) = 1 - 3 + 7 - 3 + m(1) + n = 30$
Solve the system of equations to find $m$ and $n$ .

STEP 8

To determine the number of real roots of $p(x) = 0$ , analyze the polynomial $p(x)$ and its factors.

STEP 9

Consider the degree of $p(x)$ and the nature of its factors to determine the number of real roots.
The remainder when $p(x)$ is divided by $(x-3)(x+7)$ is $R(x) = mx + n$ , and the number of real roots of $p(x) = 0$ depends on the analysis of the polynomial's factors.

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Studdy Solution

STEP 1

STEP 2

1. Use the Remainder Theorem to find conditions on the coefficients.
2. Use the given quotient information to find the remainder when dividing by $(x-3)(x+7)$ .
3. Analyze the polynomial to determine the number of real roots.

STEP 3

Using the Remainder Theorem, substitute $x = 1$ into $p(x)$ to find the remainder:
$p(1) = a(1)^3 + b(1)^2 + c(1) + 60 = 30$
Simplify:
$a + b + c + 60 = 30$ $a + b + c = -30$

STEP 4

Given that the quotient when $p(x)$ is divided 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:
$R(x) = mx + n$

STEP 5

Substitute $x = 3$ and $x = -7$ into $p(x)$ to find $m$ and $n$ :
For $x = 3$ :
$p(3) = (3-3)(3+7)(3-3) + R(3) = R(3)$ $R(3) = m(3) + n = 3m + n$
For $x = -7$ :
$p(-7) = (-7-3)(-7+7)(-7-3) + R(-7) = R(-7)$ $R(-7) = m(-7) + n = -7m + n$

STEP 6

Since $R(x)$ is the remainder when divided by $(x-3)(x+7)$ , and the quotient is $x-3$ , we know:
$p(x) = (x-3)(x+7)(x-3) + mx + n$
Given the conditions, solve for $m$ and $n$ using the system of equations from Step 3.

STEP 7

Use the fact that $p(x)$ has the remainder 30 when divided by $x-1$ to find the values of $m$ and $n$ .
$p(1) = 1 - 3 + 7 - 3 + m(1) + n = 30$
Solve the system of equations to find $m$ and $n$ .

STEP 8

To determine the number of real roots of $p(x) = 0$ , analyze the polynomial $p(x)$ and its factors.

STEP 9

Consider the degree of $p(x)$ and the nature of its factors to determine the number of real roots.
The remainder when $p(x)$ is divided by $(x-3)(x+7)$ is $R(x) = mx + n$ , and the number of real roots of $p(x) = 0$ depends on the analysis of the polynomial's factors.

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Studdy Solution

STEP 1

STEP 2

1. Use the Remainder Theorem to find conditions on the coefficients.2. Use the given quotient information to find the remainder when dividing by (x−3)(x+7) (x-3)(x+7) (x−3)(x+7).3. Analyze the polynomial to determine the number of real roots.

STEP 3

Using the Remainder Theorem, substitute x=1 x = 1 x=1 into p(x) p(x) p(x) to find the remainder:p(1)=a(1)3+b(1)2+c(1)+60=30 p(1) = a(1)^3 + b(1)^2 + c(1) + 60 = 30 p(1)=a(1)3+b(1)2+c(1)+60=30Simplify:a+b+c+60=30 a + b + c + 60 = 30 a+b+c+60=30 a+b+c=−30 a + b + c = -30 a+b+c=−30

STEP 4

STEP 5

STEP 6

STEP 7

Use the fact that p(x) p(x) p(x) has the remainder 30 when divided by x−1 x-1 x−1 to find the values of m m m and n n n.p(1)=1−3+7−3+m(1)+n=30 p(1) = 1 - 3 + 7 - 3 + m(1) + n = 30 p(1)=1−3+7−3+m(1)+n=30Solve the system of equations to find m m m and n n n.

STEP 8

To determine the number of real roots of p(x)=0 p(x) = 0 p(x)=0, analyze the polynomial p(x) p(x) p(x) and its factors.

STEP 9

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1. Use the Remainder Theorem to find conditions on the coefficients.
2. Use the given quotient information to find the remainder when dividing by $(x-3)(x+7)$ .
3. Analyze the polynomial to determine the number of real roots.

Using the Remainder Theorem, substitute $x = 1$ into $p(x)$ to find the remainder:
$p(1) = a(1)^3 + b(1)^2 + c(1) + 60 = 30$
Simplify:
$a + b + c + 60 = 30$ $a + b + c = -30$

Use the fact that $p(x)$ has the remainder 30 when divided by $x-1$ to find the values of $m$ and $n$ .
$p(1) = 1 - 3 + 7 - 3 + m(1) + n = 30$
Solve the system of equations to find $m$ and $n$ .

To determine the number of real roots of $p(x) = 0$ , analyze the polynomial $p(x)$ and its factors.