Q.Evaluate: $\int \dfrac{x^2+2}{x+1}\,dx$

We are dividing $3x^5 - 6x^4 + 2x^3 + 4x^2 - 5x + 8$ by $x^2 - 2x + 3$. The quotient will be a cubic (since degree 5 minus degree 2 gives degree 3) and the remainder a linear polynomial (degree less than 2). The problem gives the quotient as $ax^3 + bx^2 + cx + d$ and remainder as $px + q$. Our goal is to find $ab + cd$ in terms of $p$ and $q$.

Concept and intuition:

Instead of doing the full long division blindly, we can use the division algorithm:

If we expand the right-hand side and equate coefficients with the left-hand side, we get a system of equations linking $a,b,c,d,p,q$. Then we can compute $ab + cd$ and express it in terms of $p$ and $q$.

Let’s work through it step by step.

1. Set up the division identity

2. Expand the product

   First multiply $x^2$ by the quotient:

   $x^2(ax^3 + bx^2 + cx + d) = ax^5 + bx^4 + cx^3 + dx^2$.

   Then multiply $-2x$ by the quotient:

   $-2x(ax^3 + bx^2 + cx + d) = -2ax^4 - 2bx^3 - 2cx^2 - 2dx$.

   Then multiply $3$ by the quotient:

   $3(ax^3 + bx^2 + cx + d) = 3ax^3 + 3bx^2 + 3cx + 3d$.

   Summing these (and adding the remainder $px + q$):

3. Equate coefficients with the dividend

   The dividend’s coefficients are:

   $x^5: 3$, $x^4: -6$, $x^3: 2$, $x^2: 4$, $x^1: -5$, constant: $8$.

   So we have:

4. Solve for $a, b, c, d$

   From $a = 3$:

   $b - 2(3) = -6 \Rightarrow b - 6 = -6 \Rightarrow b = 0$.

   Then $c - 2(0) + 3(3) = 2 \Rightarrow c + 9 = 2 \Rightarrow c = -7$.

   Then $d - 2(-7) + 3(0) = 4 \Rightarrow d + 14 = 4 \Rightarrow d = -10$.

   So the quotient is $3x^3 + 0x^2 - 7x - 10$.

5. Find $p$ and $q$ from the last two equations

   From $-2d + 3c + p = -5$:

   $-2(-10) + 3(-7) + p = -5 \Rightarrow 20 - 21 + p = -5 \Rightarrow -1 + p = -5 \Rightarrow p = -4$.

   From $3d + q = 8$:

   $3(-10) + q = 8 \Rightarrow -30 + q = 8 \Rightarrow q = 38$.

   So remainder is $-4x + 38$.

6. Compute $ab + cd$

   $a = 3$, $b = 0$, $c = -7$, $d = -10$.

   Then $ab = 3 \cdot 0 = 0$, $cd = (-7)(-10) = 70$.

   So $ab + cd = 0 + 70 = 70$.

7. Express 70 in terms of $p$ and $q$

   We have $p = -4$, $q = 38$.

   Check each option:

   (A) $p + 2q = -4 + 76 = 72$ — no.

   (B) $p + 2q - 2 = -4 + 76 - 2 = 70$ — yes.

   (C) $2p + q = -8 + 38 = 30$ — no.

   (D) $2p + q - 2 = -8 + 38 - 2 = 28$ — no.

Thus $ab + cd = p + 2q - 2$.

Concept & Intuition

When we divide a polynomial $P(x)$ of degree 5 by a quadratic $D(x)$, the quotient has degree 3 and the remainder is linear. The problem gives us the forms of the quotient and remainder, so we perform the division to find $a,b,c,d,p,q$, then compute $ab+cd$ and match it to the options in terms of $p,q$.

Note on the dividend: the printed expression repeats an $x^4$ term ($-6x^4+2x^4$); read together with the answer choices, the intended dividend has a genuine $x^3$ term there, i.e. $P(x) = 3x^5 - 6x^4 + 2x^3 + 4x^2 - 5x + 8$ — this is the reading used below, and it is the one that reproduces an exact match with one of the given options.

Step-by-step solution

1. Set up the division.

   Dividend: $P(x) = 3x^5 - 6x^4 + 2x^3 + 4x^2 - 5x + 8$.

   Divisor: $D(x) = x^2 - 2x + 3$.

2. First term. Divide $3x^5$ by $x^2$ to get $3x^3$. Multiply: $3x^3(x^2-2x+3) = 3x^5-6x^4+9x^3$.

   Subtract: $(3x^5-6x^4+2x^3+4x^2-5x+8) - (3x^5-6x^4+9x^3) = -7x^3+4x^2-5x+8$.

3. Second term. Divide $-7x^3$ by $x^2$ to get $-7x$. Multiply: $-7x(x^2-2x+3) = -7x^3+14x^2-21x$.

   Subtract: $(-7x^3+4x^2-5x+8) - (-7x^3+14x^2-21x) = -10x^2+16x+8$.

4. Third term. Divide $-10x^2$ by $x^2$ to get $-10$. Multiply: $-10(x^2-2x+3) = -10x^2+20x-30$.

   Subtract: $(-10x^2+16x+8) - (-10x^2+20x-30) = -4x+38$.

   Since the remainder $-4x+38$ has degree less than 2, the division is complete.

5. Identify the coefficients.

   Quotient: $3x^3 + 0x^2 - 7x - 10 \;\Rightarrow\; a=3,\ b=0,\ c=-7,\ d=-10$.

   Remainder: $-4x + 38 \;\Rightarrow\; p=-4,\ q=38$.

6. Compute $ab+cd$.

7. Check the options.
   • (A) $p+2q = -4+76 = 72$
   • (B) $2p+q = -8+38 = 30$
   • (C) $2p+q-2 = 30-2 = 28$
   • (D) $p+2q-2 = 72-2 = 70$ ✓ matches

We are given the polynomial

and told that $\alpha$ is a multiple root. A multiple root means it is a root of $P(x)$ and also a root of its derivative $P'(x)$ (since the multiplicity is at least 2). The key idea: find the common roots of $P$ and $P'$ by computing their greatest common divisor (gcd). Then test each candidate in the expression $3\alpha^2 - 2\alpha + 1$.

1. Find the derivative

2. Compute the gcd of $P$ and $P'$

   We perform polynomial long division (or Euclidean algorithm). Divide $P(x)$ by $P'(x)$:

   • Leading term: $x^5 / 5x^4 = \frac{1}{5}x$. Multiply $P'(x)$ by $\frac{1}{5}x$:

• Subtract from $P(x)$:

 Simplify coefficients:

 Constant: $+8 = \frac{40}{5}$. So remainder is:

 Multiply by 5 to clear denominators (gcd is unaffected by scaling):

 Factor $-2$ (optional): $R_1(x) = -2(3x^4 - 11x^3 + 3x^2 + 24x - 20)$.

3. Continue the Euclidean algorithm

Now divide $P'(x)$ by this remainder (or its scaled version). Use $Q(x) = 3x^4 - 11x^3 + 3x^2 + 24x - 20$ for convenience.

Divide $5x^4 - 24x^3 + 33x^2 - 4x - 12$ by $3x^4 - 11x^3 + 3x^2 + 24x - 20$:

Leading term: $5x^4 / 3x^4 = \frac{5}{3}$. Multiply:

Subtract from $P'(x)$:

So remainder:

Multiply by 3:

4. Next step Divide $Q(x) = 3x^4 - 11x^3 + 3x^2 + 24x - 20$ by $R_2(x) = -17x^3 + 84x^2 - 132x + 64$. Leading term: $3x^4 / -17x^3 = -\frac{3}{17}x$. Multiply:

Subtract from $Q(x)$ (write $Q$ with denominator 17):

Subtract:

Constant: $-\frac{340}{17}$. So remainder:

Multiply by 17:

Factor 5: $R_3(x) = 5(13x^3 - 69x^2 + 120x - 68)$.

5. Continue Divide $R_2(x) = -17x^3 + 84x^2 - 132x + 64$ by $S(x) = 13x^3 - 69x^2 + 120x - 68$. Leading term: $-17/13$. Multiply:

Subtract from $R_2$ (write $R_2$ with denominator 13):

Subtract:

Constant: $\left(\frac{832}{13} - \frac{1156}{13}\right) = -\frac{324}{13}$

So remainder:

Multiply by 13: $R_4(x) = -81x^2 + 324x - 324 = -81(x^2 - 4x + 4) = -81(x-2)^2$.

6. The gcd The last nonzero remainder (up to constant factor) is $(x-2)^2$. So the common factor of $P$ and $P'$ is $x-2$. Hence the only multiple root is $\alpha = 2$.

7. Evaluate the expression