Q.Integrate the following function: $\int \frac{x^3 - x^2 + x - 1}{x - 1} dx$

Concept and Intuition

Polynomial long division by a quadratic proceeds exactly like numerical long division: at each stage, match the leading term, multiply the divisor by that matching monomial, and subtract to bring down a lower-degree remainder, repeating until the remainder's degree is less than the divisor's.

Step-by-Step Solution

1. $3x^5\div x^2=3x^3$. Subtract $3x^3(x^2+x-3)=3x^5+3x^4-9x^3$ from the dividend: remainder $-7x^4+14x^3-3x^2+6x-8$.
2. $-7x^4\div x^2=-7x^2$. Subtract $-7x^2(x^2+x-3)=-7x^4-7x^3+21x^2$: remainder $21x^3-24x^2+6x-8$.
3. $21x^3\div x^2=21x$. Subtract $21x(x^2+x-3)=21x^3+21x^2-63x$: remainder $-45x^2+69x-8$.
4. $-45x^2\div x^2=-45$. Subtract $-45(x^2+x-3)=-45x^2-45x+135$: remainder $114x-143$ (degree $<2$, so division stops). …

We are told that when

is divided by $x^2 + x + 1$, the remainder is $5$. That means

where $Q(x)$ is a cubic polynomial (since dividing a degree‑5 polynomial by a degree‑2 polynomial gives a degree‑3 quotient). Our job is to find $Q(x)$ from the options.

Why polynomial long division works here

If we actually performed long division, we’d subtract multiples of the divisor until the remainder’s degree is less than 2. But we can be smarter: since the remainder is a constant, the division equation must hold identically for all $x$. That means the coefficients of like powers of $x$ on both sides must match. We can use this to solve for both the unknown $k$ and the unknown quotient coefficients.

Step-by-step solution

1. Set up the unknown quotient Let

where $a, b, c, d$ are real numbers we need to find.

2. Write the division identity

3. Expand the product Multiply term by term:

4. Collect like terms

So the product is:

5. Add the remainder 5 The right side becomes:

6. Match coefficients with the left side $$$$

Concept and Intuition

If $x^2-x-1$ (degree 2) divides $ax^3+bx^2+0x+1$ (degree 3) exactly, the quotient must be linear, say $ax+c$. Expanding $(x^2-x-1)(ax+c)$ and matching every coefficient with the target cubic pins down both $a$ (already known from the leading term) and the unknowns $b,c$.

Step-by-Step Solution

1. Let the quotient be $ax+c$ (leading coefficient must be $a$ to match $x^3$ coefficient). Then:

2. Match to $ax^3+bx^2+0\cdot x+1$:

• $x^2$: $c-a=b$
• $x^1$: $-(a+c)=0 \Rightarrow c=-a$
• $x^0$: $-c=1 \Rightarrow c=-1$

3. From $c=-1$ and $c=-a$: $-1=-a \Rightarrow a=1$.
4. From $c-a=b$: $b=-1-1=-2$. …

Concept and Intuition

If $x^2+px+1$ divides $ax^3+bx+c$ exactly, the quotient must be linear (degree $3-2=1$), say $Ax+B$. Multiplying out and comparing coefficients on both sides — including the "missing" $x^2$ term (coefficient 0) on the left — pins down $A,B,p$ in terms of $a,b,c$, and eliminating $p$ gives the required relation.

Step-by-Step Solution

1. Write $ax^3+0\cdot x^2+bx+c=(x^2+px+1)(Ax+B)$.
2. Expand the right side: $Ax^3+(B+Ap)x^2+(Bp+A)x+B$.
3. Match coefficients:
   • $x^3$: $A=a$.
   • $x^2$: $B+Ap=0\Rightarrow B=-ap$.
   • constant: $B=c\Rightarrow c=-ap\Rightarrow p=-c/a$.
   • $x^1$: $Bp+A=b\Rightarrow (-ap)p+a=b\Rightarrow a(1-p^2)=b$. …

Concept and Intuition

If $x^2+px+1$ exactly divides the cubic $ax^3+0x^2+bx+c$, the quotient must be linear, $ax+q$ (leading coefficient $a$ to match $ax^3$). Multiplying out and matching each power of $x$ gives three equations in $p,q$ that must be consistent — eliminating them gives the required relation among $a,b,c$.

Step-by-Step Solution

1. Assume $ax^3+bx+c=(x^2+px+1)(ax+q)$.
2. Expand RHS: $ax^3+qx^2+apx^2+pqx+ax+q = ax^3+(q+ap)x^2+(pq+a)x+q$.
3. Match $x^2$: $q+ap=0 \Rightarrow q=-ap$.
4. Match $x$: $pq+a=b \Rightarrow p(-ap)+a=b \Rightarrow -ap^2=b-a \Rightarrow p^2=\dfrac{a-b}{a}$.
5. Match constant: $q=c \Rightarrow -ap=c \Rightarrow p=-\dfrac{c}{a} \Rightarrow p^2=\dfrac{c^2}{a^2}$. …

Concept and Intuition

Dividing by $x^2-x-3$ is equivalent to working modulo the relation $x^2=x+3$. Repeatedly substituting this relation reduces any higher power of $x$ down to a linear expression $ax+b$, which is exactly the remainder — no need for a full long-division table.

Step-by-Step Solution

1. From $x^2-x-3=0$, we get $x^2=x+3$.
2. $x^3=x\cdot x^2=x(x+3)=x^2+3x=(x+3)+3x=4x+3$.
3. $x^4=x\cdot x^3=x(4x+3)=4x^2+3x=4(x+3)+3x=7x+12$.
4. $x^5=x\cdot x^4=x(7x+12)=7x^2+12x=7(x+3)+12x=19x+21$.
5. Substitute into the polynomial: $2x^5=38x+42$, $-3x^4=-21x-36$, $5x^3=20x+15$, $-3x^2=-3x-9$, $7x=7x$, $-9=-9$. …

Concept and Intuition

By the Remainder Theorem, the remainder when $f(x)$ is divided by $(x-c)$ is simply $f(c)$. This turns the problem into solving a small linear system for the unknown coefficients $a,b$.

Step-by-Step Solution

1. $f(x)=x^4-2x^3+3x^2-ax+b$.
2. $f(1)=1-2+3-a+b=2-a+b=5 \Rightarrow b-a=3$.
3. $f(-1)=1+2+3+a+b=6+a+b=19 \Rightarrow a+b=13$.
4. Adding the two equations ($b-a=3$ and $a+b=13$): $2b=16\Rightarrow b=8$, then $a=5$. …

Concept and Intuition

For a quartic $x^4+px^3+qx^2+rx+s=0$ with roots $\alpha,\beta,\gamma,\delta$: sum of roots $=-p$, sum of pairwise products $=q$, sum of triple products $=-r$, product $=s$. Given a symmetric constraint like $\alpha+\beta=0$, substituting $\beta=-\alpha$ into these symmetric sums lets many cross terms vanish, isolating $\alpha^2$ and then $\gamma,\delta$.

Step-by-Step Solution

1. Here $p=-1,q=-8,r=2,s=12$, so:

• $\alpha+\beta+\gamma+\delta = 1$
• $\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta=-8$
• $\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta=-2$
• $\alpha\beta\gamma\delta=12$

2. Given $\alpha+\beta=0$, so $\beta=-\alpha$, and hence $\gamma+\delta=1$.
3. Triple-product sum: $\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta = \alpha\beta(\gamma+\delta) + \gamma\delta(\alpha+\beta) = \alpha\beta(1) + \gamma\delta(0) = \alpha\beta$. With $\alpha\beta=-\alpha^2$ (since $\beta=-\alpha$): $-\alpha^2 = -2\Rightarrow \alpha^2=2$. …

Concept and Intuition

If a quartic $f(x)$ is known to have $x^2-5x+6$ as one of its two quadratic factors, then $f(x)=(x^2-5x+6)(x^2+px+q)$ for some unknowns $p,q$. Multiplying out and comparing coefficients of $f(x)=x^4-17x^3+kx^2-247x+210$ term-by-term pins down $p$ and $q$ directly — we don't even need to know $k$, since we can use the $x^3$ and constant coefficients (which don't involve $k$) and then verify with the $x$-coefficient.

Step-by-Step Solution

1. Let $f(x)=(x^2-5x+6)(x^2+px+q)$.
2. Expand: $x^4+px^3+qx^2-5x^3-5px^2-5qx+6x^2+6px+6q$ $=x^4+(p-5)x^3+(q-5p+6)x^2+(6p-5q)x+6q$.
3. Match $x^3$ coefficient: $p-5=-17\Rightarrow p=-12$.
4. Match constant term: $6q=210\Rightarrow q=35$. …

Find $\alpha$ (Remainder Theorem). Divide $x^5-2x^4+3x^3-4x^2-x+2$ by $(x-2)$: evaluate at $x=2$.

Confirm $\alpha=8$ is a root of the quartic $x^4-6x^3-35x^2+132x+160$:

Remove the known root. Synthetic division of the quartic by $(x-8)$ gives

whose roots are the other three. Here $e_1=-2,\ e_2=-19,\ e_3=20$. …

Concept and Intuition

If a polynomial $Q(x)$ divides $P(x)$, every root of $Q$ must also be a root of $P$ (the Factor/Remainder Theorem). Here $x^2+x-6$ factors easily, giving two concrete numbers to plug into the cubic, turning the divisibility condition into two linear equations in the unknowns $a,b$.

Step-by-Step Solution

1. Factor: $x^2+x-6=(x+3)(x-2)$, so roots are $x=-3$ and $x=2$.
2. Since $x^2+x-6\mid 2x^3+x^2+ax+b$, both $x=2$ and $x=-3$ must satisfy $2x^3+x^2+ax+b=0$.
3. At $x=2$: $2(8)+4+2a+b=0\Rightarrow 20+2a+b=0\Rightarrow 2a+b=-20$.
4. At $x=-3$: $2(-27)+9-3a+b=0\Rightarrow-45-3a+b=0\Rightarrow-3a+b=45$. …

Concept and Intuition

Substituting $x=-y$ into $x^m+y^m$ tests whether $(x+y)$ is a factor: if $x=-y$ makes the expression vanish, then $(x+y)$ divides it (factor theorem). This substitution behaves differently depending on the parity of $m$, which is exactly why the divisibility depends on whether $m$ is odd or even.

Step-by-Step Solution

1. By the Factor Theorem, $(x+y)$ divides a polynomial $P(x,y)$ (in $x$, treating $y$ as a constant) if and only if setting $x=-y$ makes $P(-y,y)=0$.
2. Compute $P(-y,y) = (-y)^m + y^m$.
3. If $m$ is odd: $(-y)^m = -y^m$, so $P(-y,y) = -y^m+y^m = 0$. Hence $(x+y)$ divides $x^m+y^m$ for all odd $m$.
4. If $m$ is even: $(-y)^m = y^m$, so $P(-y,y) = y^m+y^m = 2y^m \ne 0$ in general. Hence $(x+y)$ does not divide $x^m+y^m$ for even $m$. …