Question

1. divide the following polynomials.

$(5c^{2} + 24c + 25) \div (c + 3)$
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1. use the remainder theorem to find the remainder of the quotient:

$\frac{k^{3}+k^{2}-4k + 2}{(k + 2)}$
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1. use the remainder theorem to find the remainder of the quotient:

$\frac{-6y^{4}+5y^{3}-4y^{2}-y + 2}{(y + 1)}$

Explanation:

Question 11

Step1: Use polynomial long division

Divide the leading term of the dividend \(5c^2 + 24c + 25\) by the leading term of the divisor \(c + 3\). So, \(\frac{5c^2}{c}=5c\).
Multiply the divisor \(c + 3\) by \(5c\): \(5c(c + 3)=5c^2 + 15c\).
Subtract this from the dividend: \((5c^2 + 24c + 25)-(5c^2 + 15c)=9c + 25\).

Step2: Divide the new leading term

Now, divide the leading term of \(9c + 25\) (which is \(9c\)) by the leading term of the divisor \(c\). So, \(\frac{9c}{c}=9\).
Multiply the divisor \(c + 3\) by \(9\): \(9(c + 3)=9c + 27\).
Subtract this from \(9c + 25\): \((9c + 25)-(9c + 27)= - 2\).

Step1: Recall the Remainder Theorem

The Remainder Theorem states that if a polynomial \(f(k)\) is divided by \(k - a\), the remainder is \(f(a)\). In this case, the divisor is \(k + 2=k-(-2)\), so \(a=-2\).

Step2: Evaluate the polynomial at \(k = - 2\)

Let \(f(k)=k^3 + k^2-4k + 2\). Substitute \(k=-2\) into \(f(k)\):
\[

$$\begin{align*} f(-2)&=(-2)^3+(-2)^2-4(-2)+2\\ &=-8 + 4+8 + 2\\ &=(-8 + 4)+(8 + 2)\\ &=-4 + 10\\ &=6 \end{align*}$$

\]

Step1: Recall the Remainder Theorem

The Remainder Theorem states that if a polynomial \(f(y)\) is divided by \(y - a\), the remainder is \(f(a)\). Here, the divisor is \(y + 1=y-(-1)\), so \(a = - 1\).

Step2: Evaluate the polynomial at \(y=-1\)

Let \(f(y)=-6y^4+5y^3-4y^2-y + 2\). Substitute \(y = - 1\) into \(f(y)\):
\[

$$\begin{align*} f(-1)&=-6(-1)^4+5(-1)^3-4(-1)^2-(-1)+2\\ &=-6(1)+5(-1)-4(1)+1 + 2\\ &=-6-5-4 + 1+2\\ &=(-6-5-4)+(1 + 2)\\ &=-15 + 3\\ &=-12 \end{align*}$$

\]

Answer:

The quotient is \(5c + 9\) and the remainder is \(-2\), so we can write it as \(5c + 9-\frac{2}{c + 3}\)

Question 12