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The expression
A(x-1)^(3)+B(x+3)^(2)+20 is exactly divisible by
x+1 and leaves a remainder of 26 when divided by
x. Find the value of
A and of
B. Using these value found, rewrite the expression as a polynomial and factorize it completel

$4$ . The expression $A(x-1)^{3}+B(x+3)^{2}+20$ is exactly divisible by $x+1$ and leaves a remainder of $26$ when divided by $x$ . Find the value of $A$ and of $B$ . Using these value found, rewrite the expression as a polynomial and factorize it completel

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Write a polynomial from its roots

Full solution

Q.

4

. The expression

A(x-1)^{3}+B(x+3)^{2}+20

is exactly divisible by

x+1

and leaves a remainder of

26

when divided by

x

. Find the value of

A

and of

B

. Using these value found, rewrite the expression as a polynomial and factorize it completel

Substitute $x = -1$ : Since the expression is exactly divisible by $x+1$ , the remainder when divided by $x+1$ is $0$ . Let's substitute $x = -1$ into the expression to find $A$ .
$A(-1-1)^{3}+B(-1+3)^{2}+20 = 0$
$A(-2)^{3}+B(2)^{2}+20 = 0$
$-8A+4B+20 = 0$
Solve for $A$ in terms of $B$ : Now let's solve for $A$ in terms of $B$ . $-8A = -4B - 20$ $A = \frac{4B + 20}{8}$ $A = \frac{B}{2} + \frac{5}{2}$
Set constant term equal to $26$ : Next, since the expression leaves a remainder of $26$ when divided by $x$ , the constant term must be $26$ . Let's set the constant term equal to $26$ and solve for $B$ .
$A(-1)^{3}+B(3)^{2}+20 = 26$
$-A+9B+20 = 26$
$-A = 26 - 9B - 20$
$-A = 6 - 9B$
Substitute A into equation: Substitute $A$ from the second step into this equation. $\newline$ $\frac{B}{2} + \frac{5}{2} = 6 - 9B$ $\newline$ $B + 5 = 12 - 18B$ $\newline$ $19B = 7$ $\newline$ $B = \frac{7}{19}$
Solve for B: Now substitute B back into the equation for A. $\newline$ $A = \frac{B}{2} + \frac{5}{2}$ $\newline$ $A = \left(\frac{7}{19}\right)/2 + \frac{5}{2}$ $\newline$ $A = \frac{7}{38} + \frac{95}{38}$ $\newline$ $A = \frac{102}{38}$ $\newline$ $A = \frac{17}{6}$
Substitute $B$ into $A$ equation: Now that we have $A$ and $B$ , let's rewrite the expression. $A(x-1)^{3}+B(x+3)^{2}+20$ $\left(\frac{17}{6}\right)(x-1)^{3}+\left(\frac{7}{19}\right)(x+3)^{2}+20$
Rewrite the expression: Finally, let's factorize the polynomial completely. Since we don't have a common factor for all terms, we can't factorize it further without expanding the expression. However, the question asked for the values of $A$ and $B$ , which we have found.

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