IB exam questions: topic 1

We expand using $(a+b{)}^3=a^3+3a^{2}b+3a{b}^2+b^3$, with $a=\sqrt{3}$, $b=-2$: $$(\sqrt{3}-2{)}^3=(\sqrt{3}{)}^3+3(\sqrt{3}{)}^2(-2)+3\sqrt{3}(-2{)}^2+(-2{)}^3.$$ We calculate each term:

• $(\sqrt{3}{)}^3=3\sqrt{3}$
• $3\cdot 3\cdot (-2)=-18$
• $3\sqrt{3}\cdot 4=12\sqrt{3}$
• $(-2{)}^3=-8$

Adding: $3\sqrt{3}+12\sqrt{3}-18-8=15\sqrt{3}-26$.

Answer: $15\sqrt{3}-26$ (that is, $a=15$, $b=-26$)

1. By the remainder theorem, the remainder when dividing by $(x+1)$ is $f(-1)=7$: $$1-p-q+5=7\Rightarrow p+q=-1.$$
2. The remainder when dividing by $(x-2)$ is $f(2)=1$: $$16+8p+2q+5=1\Rightarrow 8p+2q=-20\Rightarrow 4p+q=-10.$$
3. Subtracting the first equation from the second: $3p=-9$, so $p=-3$.
4. Substituting into $p+q=-1$: $q=-1+3=2$.

Answer: $p=-3,q=2$

1. Let $f(x)=3x^3+ax+b$. By the remainder theorem, $f(2)=2$: $$24+2a+b=2\Rightarrow 2a+b=-22.$$
2. Also $f(-1)=5$: $$-3-a+b=5\Rightarrow -a+b=8.$$
3. Subtracting the second equation from the first: $3a=-30$, so $a=-10$.
4. Substituting into $-a+b=8$: $10+b=8$, so $b=-2$.
5. Check in the first: $2(-10)+(-2)=-22$. Correct.

Answer: $a=-10,b=-2$

1. If $x=k$ is a solution, we substitute: $k^3+k\cdot k^2-k-k=0$.
2. We simplify: $2k^3-2k=0$.
3. Factor out the common factor: $2k(k^2-1)=2k(k-1)(k+1)=0$.
4. Set each factor equal to zero: $k=0$, $k=1$, $k=-1$.

Answer: $k=-1,0,1$

a) We substitute $x=-2$: $P(-2)=(-2{)}^3-3(-2{)}^2+4(-2)+c=-8-12-8+c=c-28$.

b) By the remainder theorem, the remainder when dividing by $(x+2)$ is $P(-2)$. Then $c-28=-23$, so $c=5$.

Answer: a) $P(-2)=c-28$; b) $c=5$

1. By the remainder theorem, the remainder when $f(x)$ is divided by $(x+1)$ is $f(-1)$, so $f(-1)=-20$.

2. We substitute: $f(-1)=6(-1{)}^4+11(-1{)}^3-22(-1{)}^2+a(-1)+6=6-11-22-a+6=-21-a$.

3. Equate: $-21-a=-20\Rightarrow a=-1$.

Answer: $a=-1$

1. By the factor theorem, $(x+2)$ is a factor of $f(x)$ if and only if $f(-2)=0$.
2. We calculate: $$f(-2)=(-2{)}^3-2(-2{)}^2-5(-2)+k=-8-8+10+k=k-6$$
3. Set equal to zero: $k-6=0$.

Answer: $k=6$

1. By the remainder theorem, the remainder when dividing by $(x-1)$ is the value of the polynomial at $x=1$.
2. We substitute: $$1^4+a\cdot 1+3=a+4$$
3. Equate to the given remainder: $a+4=8$.
4. Therefore $a=4$.

Answer: $a=4$

1. If $P(x)$ is divisible by $x+1$, the remainder theorem gives $P(-1)=0$: $$-1+a-b+2=0\Rightarrow a-b=-1.$$
2. If it is divisible by $x-2$, then $P(2)=0$: $$8+4a+2b+2=0\Rightarrow 2a+b=-5.$$
3. Adding the two equations: $3a=-6$, so $a=-2$.
4. From $a-b=-1$: $b=a+1=-1$.
5. Check: $2(-2)+(-1)=-5$ holds.

Answer: $a=-2,b=-1$

1. Remainder theorem: the remainder when dividing by $x-2$ is $f(2)=15$: $$16+24+4p-4+q=15\Rightarrow 4p+q=-21.$$
2. If $x+3$ is a factor, then $f(-3)=0$: $$81-81+9p+6+q=0\Rightarrow 9p+q=-6.$$
3. Subtracting the first equation from the second: $5p=15$, so $p=3$.
4. Substituting: $q=-21-4(3)=-33$.
5. Check: $9(3)+(-33)=-6$ holds.

Answer: $p=3,q=-33$

1. Remainder when dividing by $x+1$ (remainder theorem): $q(-1)=-1+k+7+3=k+9$.
2. Remainder when dividing by $x+2$: $q(-2)=-8+4k+14+3=4k+9$.
3. The first remainder is seven times the second: $$k+9=7(4k+9)=28k+63.$$
4. Solving: $-27k=54$, so $k=-2$.
5. Check: the remainders are $q(-1)=7$ and $q(-2)=1$, and $7=7\cdot 1$.

Answer: $k=-2$

Let $f(x)=3x^5-ax+b$. By the remainder theorem, the remainders are $f(1)$ and $f(-1)$.

a) 1. $f(1)=3-a+b$ and $f(-1)=-3+a+b$. 2. Equating the remainders: $3-a+b=-3+a+b$. 3. The $b$ terms cancel: $6=2a$, so $a=3$.

b) 1. With $a=3$ both remainders equal $b$ ($f(1)=b$ and $f(-1)=b$), and the condition holds for any $b$. 2. As the value of $b$ is not restricted, $b$ can be any real number.

Answer: a) $a=3$; b) $b\in \mathbb{R}$

By the remainder theorem, the remainder when a polynomial $f(x)$ is divided by $x+1$ is $f(-1)$.

1. First polynomial: $f(-1)=2(-1{)}^3+k(-1{)}^2+6(-1)+32=-2+k-6+32=k+24$.
2. Second polynomial: $g(-1)=(-1{)}^4-6(-1{)}^2-k^2(-1)+9=1-6+k^2+9=k^2+4$.
3. The remainders are equal: $$k^2+4=k+24⟹k^2-k-20=0⟹(k-5)(k+4)=0$$
4. Therefore $k=5$ or $k=-4$.

Answer: $k=5$ or $k=-4$

Let $f(x)=3x^3+p{x}^2+qx-2$.

1. Since $(x+2)$ is a divisor, by the factor theorem $f(-2)=0$: $$3(-8)+4p-2q-2=0⟹4p-2q=26⟹2p-q=13$$
2. The remainder when dividing by $(x+1)$ is $f(-1)=4$: $$-3+p-q-2=4⟹p-q=9$$
3. Subtracting the second equation from the first: $(2p-q)-(p-q)=13-9\Rightarrow p=4$.
4. From $p-q=9$: $q=p-9=-5$.
5. Check: $f(-2)=-24+16+10-2=0$ and $f(-1)=-3+4+5-2=4$. Correct.

Answer: $p=4$, $q=-5$

1. By the remainder theorem, the remainder when $P(x)$ is divided by $x-a$ is $P(a)$. We require it to equal 2: $$a^2+4a-b=2\Rightarrow b=a^2+4a-2.$$
2. Completing the square: $$b=(a+2{)}^2-4-2=(a+2{)}^2-6.$$
3. Since $(a+2{)}^2\ge 0$ for all $a\in \mathbb{R}$, the minimum is reached when $a=-2$, and then $b=-6$.

Answer: $b_{\min }=-6$

1. We factorise the divisor: $x^2-4x+3=(x-1)(x-3)$.
2. If it is a divisor of $P(x)=x^3+(a-4)x^2+(3-4a)x+3$, then $P(1)=0$ and $P(3)=0$ (factor theorem).
3. With $x=1$: $$1+(a-4)+(3-4a)+3=0\Rightarrow 3-3a=0\Rightarrow a=1.$$
4. Check with $x=3$: $$27+9(a-4)+3(3-4a)+3=27+9a-36+9-12a+3=3-3a=0\Rightarrow a=1.$$ Both conditions give the same value.

Answer: $a=1$

1. Remainder theorem with $(x-1)$: $P(1)=-2$. $$4+p+q+1=-2\Rightarrow p+q=-7.(1)$$
2. With $(2x-1)$ the remainder is $P\left(\frac{1}{2}\right)=\frac{13}{4}$: $$4\cdot \frac{1}{8}+\frac{p}{4}+\frac{q}{2}+1=\frac{13}{4}\Rightarrow \frac{1}{2}+1+\frac{p}{4}+\frac{q}{2}=\frac{13}{4}.$$ So $\frac{p}{4}+\frac{q}{2}=\frac{13}{4}-\frac{3}{2}=\frac{7}{4}$. Multiplying by 4: $$p+2q=7.(2)$$
3. Subtracting (1) from (2): $q=14$. From (1): $p=-7-14=-21$.

Answer: $p=-21,q=14$

1. By the factor theorem, $f(2)=0$ and $f(-2)=0$.
2. $f(2)=8+4p+2q+4=0\Rightarrow 2p+q=-6$. (1)
3. $f(-2)=-8+4p-2q+4=4p-2q-4=0\Rightarrow 2p-q=2$. (2)
4. Adding (1) and (2): $4p=-4$, so $p=-1$.
5. From (1): $q=-6-2p=-6+2=-4$.
6. Check: $f(x)=x^3-x^2-4x+4$; $f(2)=8-4-8+4=0$ and $f(-2)=-8-4+8+4=0$.

Answer: $p=-1,q=-4$

a) 1. By the remainder theorem, the remainder when $f(x)$ is divided by $x-a$ is $f(a)$. So $f(a)=-10$.

2. Substituting: $4a^3+2a\cdot a-7a=-10$, that is, $4a^3+2a^2-7a+10=0$.

3. Testing integer values: for $a=-2$, $4(-8)+2(4)-7(-2)+10=-32+8+14+10=0$. So $(a+2)$ is a factor.

4. Dividing: $4a^3+2a^2-7a+10=(a+2)(4a^2-6a+5)$.

5. The quadratic factor has discriminant $\Delta =(-6{)}^2-4\cdot 4\cdot 5=36-80=-44<0$, so it has no real roots. The only real solution is $a=-2$.

b) 1. With $a=-2$: $f(x)=4x^3-4x+14$.

2. Differentiating: $f^{\prime }(x)=12x^2-4=0\Rightarrow x=\pm \frac{1}{\sqrt{3}}$.

3. Value at the local maximum ($x=-\frac{1}{\sqrt{3}}$): $f=4\left(-\frac{1}{3\sqrt{3}}\right)+\frac{4}{\sqrt{3}}+14=\frac{8}{3\sqrt{3}}+14>0$.

4. Value at the local minimum ($x=\frac{1}{\sqrt{3}}$): $f=\frac{4}{3\sqrt{3}}-\frac{4}{\sqrt{3}}+14=14-\frac{8}{3\sqrt{3}}\approx 12.5>0$.

5. Both extreme values are positive, and since $f(x)\to -\infty$ as $x\to -\infty$, the graph cuts the $x$-axis only once (to the left of the local maximum). Therefore $f(x)=0$ has a unique real solution.

6. Using the calculator, the root is $x\approx -1.74$ (check: $4(-1.74{)}^3-4(-1.74)+14\approx -0.1\approx 0$).

Answer: a) $a=-2$; b) $x=-1.74$

1. Moving everything to one side: $5x^3+48x^2+100x+(2-a)=0$.

2. By Vieta's formulas for a cubic $p{x}^3+q{x}^2+rx+s=0$:

$r_1+r_2+r_3=-\frac{q}{p}=-\frac{48}{5}$, $r_1{r}_2{r}_3=-\frac{s}{p}=-\frac{2-a}{5}=\frac{a-2}{5}$.

3. Imposing the given condition: $-\frac{48}{5}+\frac{a-2}{5}=0\Rightarrow a-2=48$.

Answer: $a=50$

1. If it is divisible by $(x-1)$, $(x-2)$ and $(x-3)$, then $1$, $2$ and $3$ are roots (factor theorem).
2. A polynomial of degree 4 has four roots. Let $s$ be the fourth root. The product of the roots of a monic polynomial of degree 4 is the constant term: $1\cdot 2\cdot 3\cdot s=6$, so $s=1$.
3. Therefore $$x^4+p{x}^3+q{x}^2+rx+6=(x-1{)}^2(x-2)(x-3).$$
4. We expand: $(x-2)(x-3)=x^2-5x+6$ and $(x-1{)}^2=x^2-2x+1$. Then $$(x^2-2x+1)(x^2-5x+6)=x^4-5x^3+6x^2-2x^3+10x^2-12x+x^2-5x+6$$ $$=x^4-7x^3+17x^2-17x+6.$$
5. Compare coefficients.

Answer: $p=-7,q=17,r=-17$

a) If $(x-4)$ is a factor, then by the factor theorem $q(4)=0$: $$q(4)=3\cdot 64-11\cdot 16+4k+8=192-176+4k+8=24+4k=0\Rightarrow k=-6.$$

b) Now $q(x)=3x^3-11x^2-6x+8$. We divide by $(x-4)$: we seek $q(x)=(x-4)(3x^2+bx+c)$. Comparing coefficients: $3x^3$ matches; $b-12=-11\Rightarrow b=1$; $c-4b=-6\Rightarrow c=-2$; $-4c=8$ holds. So $$q(x)=(x-4)(3x^2+x-2).$$ We factorise the quadratic: $3x^2+x-2=(3x-2)(x+1)$ (check: $3x^2+3x-2x-2$).

Answer: a) $k=-6$; b) $q(x)=(x-4)(x+1)(3x-2)$

1. Remainder theorem: $p(-1)=-1$ and $p(2)=27$.

2. $p(-1)=(-a+b{)}^3=-1$. As the real cube root of $-1$ is $-1$ (and $a,b$ are real): $b-a=-1$.

3. $p(2)=(2a+b{)}^3=27\Rightarrow 2a+b=3$.

4. We solve the system $\{\begin{array}{l}b-a=-1\\ 2a+b=3\end{array}$. Subtracting the first equation from the second: $3a=4$, so $a=\frac{4}{3}$.

5. Then $b=a-1=\frac{4}{3}-1=\frac{1}{3}$.

Answer: $a=\frac{4}{3},b=\frac{1}{3}$

1. By the remainder theorem, the remainder when dividing by $(x-2)$ is $f(2)$, and the remainder when dividing by $(x+1)$ is $f(-1)$.
2. We calculate each one: $$f(2)=8+12+2a+b=20+2a+b$$ $$f(-1)=-1+3-a+b=2-a+b$$
3. Since the remainders are equal: $$20+2a+b=2-a+b$$
4. Cancel $b$ and group terms: $3a=-18$.
5. Therefore $a=-6$.

Answer: $a=-6$

Let $p(x)=x^3+a{x}^2-3x+b$.

1. Divisible by $(x-2)$ means $p(2)=0$: $$8+4a-6+b=0\Rightarrow 4a+b=-2$$
2. The remainder when dividing by $(x+1)$ is $p(-1)=6$: $$-1+a+3+b=6\Rightarrow a+b=4$$
3. Subtract the second equation from the first: $3a=-6$, so $a=-2$.
4. From $a+b=4$ we obtain $b=6$.
5. Check: $4(-2)+6=-2$. Correct.

Answer: $a=-2,b=6$

Let $p(x)=x^3-2x^2+ax+b$.

1. Having the factor $(x-1)$ means $p(1)=0$: $$1-2+a+b=0\Rightarrow a+b=1$$
2. The remainder when dividing by $(x+1)$ is $p(-1)=8$: $$-1-2-a+b=8\Rightarrow -a+b=11$$
3. Add both equations: $2b=12$, so $b=6$.
4. From $a+b=1$ we obtain $a=-5$.
5. Check: $-(-5)+6=11$. Correct.

Answer: $a=-5,b=6$

1. By the factor theorem, if $P(x)$ is divisible by $(x-1)$, then $P(1)=0$: $$2+a-4+b=0\Rightarrow a+b=2.$$
2. If it is divisible by $(x+3)$, then $P(-3)=0$: $$2(-27)+9a+12+b=0\Rightarrow 9a+b=42.$$
3. Subtract the first equation from the second: $8a=40$, so $a=5$.
4. Substituting: $b=2-5=-3$.

Answer: $a=5$, $b=-3$

Let $f(x)=A{x}^3+B{x}^2+x+6$. Being divisible by $(x+1)(x-2)$ implies $f(-1)=0$ and $f(2)=0$ (factor theorem).

1. $f(-1)=-A+B-1+6=0\Rightarrow B-A=-5$.
2. $f(2)=8A+4B+2+6=0\Rightarrow 8A+4B=-8\Rightarrow 2A+B=-2$.
3. Subtract equation 1 from equation 2: $(2A+B)-(B-A)=3A=-2-(-5)=3$, so $A=1$.
4. From equation 1: $B=A-5=-4$.

Answer: $A=1$, $B=-4$

1. By the remainder theorem, the remainder when $p(x)$ is divided by $(x-a)$ is $p(a)$. So $p(a)=-7$: $$3a^3+a\cdot a+5a=-7\Rightarrow 3a^3+a^2+5a+7=0$$
2. We test $a=-1$: $3(-1)+1-5+7=0$. So $a=-1$ is a solution and $(a+1)$ is a factor.
3. We divide (or factorise): $3a^3+a^2+5a+7=(a+1)(3a^2-2a+7)$. Check: $3a^3-2a^2+7a+3a^2-2a+7=3a^3+a^2+5a+7$.
4. For the quadratic factor $3a^2-2a+7=0$ the discriminant is $\Delta =(-2{)}^2-4\cdot 3\cdot 7=4-84=-80<0$, so it has no real roots.
5. Therefore the only real solution is $a=-1$.

Answer: $a=-1$ (the other two values of $a$ are not real numbers)

1. We write the equation as $x^3-k{x}^2+0x+3k=0$. The relations of Vieta with roots $\alpha ,\beta ,\alpha +\beta$ give:

• sum of the roots: $\alpha +\beta +(\alpha +\beta )=k\Rightarrow \alpha +\beta =\frac{k}{2}$;
• product of the three roots: $\alpha \beta (\alpha +\beta )=-3k$.

2. We substitute $\alpha \beta =-\frac{k^2}{4}$ and $\alpha +\beta =\frac{k}{2}$ into the product: $$-\frac{k^2}{4}\cdot \frac{k}{2}=-3k\Rightarrow \frac{k^3}{8}=3k.$$
3. Since $k>0$, we divide by $k$: $k^2=24$, and as it is positive, $k=\sqrt{24}=2\sqrt{6}$.
4. Check with the sum of the products in pairs (which must equal $0$): $\alpha \beta +(\alpha +\beta {)}^2=-\frac{k^2}{4}+\frac{k^2}{4}=0$. It is consistent.

Answer: $k=2\sqrt{6}$

1. Factorise the denominator: $x^2-x=x(x-1)$.

2. Set up the partial fractions: $\frac{3x-4}{x(x-1)}=\frac{A}{x}+\frac{B}{x-1}$.

3. Multiply by $x(x-1)$: $3x-4=A(x-1)+Bx$.

4. Substitute values of $x$:

• $x=0$: $-4=-A\Rightarrow A=4$.
• $x=1$: $-1=B$.

5. Therefore $\frac{3x-4}{x^2-x}=\frac{4}{x}-\frac{1}{x-1}$.

Answer: $\frac{3x-4}{x^2-x}=\frac{4}{x}-\frac{1}{x-1}$

1. We factorise the denominator: $x^4+x^3-2x^2=x^2(x^2+x-2)=x^2(x+2)(x-1)$.

2. We set up:

$$\frac{12}{x^2(x+2)(x-1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+2}+\frac{D}{x-1}$$

3. We multiply by the denominator: $12=Ax(x+2)(x-1)+B(x+2)(x-1)+C{x}^2(x-1)+D{x}^2(x+2)$.

4. We substitute convenient values:

• $x=0$: $12=B(2)(-1)\Rightarrow B=-6$.
• $x=-2$: $12=C(4)(-3)\Rightarrow C=-1$.
• $x=1$: $12=D(1)(3)\Rightarrow D=4$.

5. For $A$ we compare the coefficients of $x^3$: $0=A+C+D\Rightarrow A=-(-1)-4=-3$.

6. Check with $x=-1$: the left-hand side is $\frac{12}{1\cdot 1\cdot (-2)}=-6$ and the right-hand side is $3-6-1-2=-6$. Correct.

Answer: $\frac{12}{x^4+x^3-2x^2}=-\frac{3}{x}-\frac{6}{x^2}-\frac{1}{x+2}+\frac{4}{x-1}$

1. The denominator has a linear factor and an irreducible quadratic factor ($3x^2+4$ has no real roots), so the decomposition is $$\frac{11x^2-9x+19}{(x-1)(3x^2+4)}=\frac{A}{x-1}+\frac{Bx+C}{3x^2+4}.$$
2. We multiply by the common denominator: $$11x^2-9x+19=A(3x^2+4)+(Bx+C)(x-1).$$
3. With $x=1$: $11-9+19=21=A\cdot 7$, so $A=3$.
4. We expand with $A=3$: $11x^2-9x+19=9x^2+12+B{x}^2+(C-B)x-C$.
5. Equating coefficients of $x^2$: $9+B=11$, so $B=2$.
6. Equating constant terms: $12-C=19$, so $C=-7$.
7. Check with the coefficient of $x$: $C-B=-7-2=-9$. This agrees.

Answer: $\frac{3}{x-1}+\frac{2x-7}{3x^2+4}$

For $2x^2+4x-1=0$, by Vieta's formulas: $\alpha +\beta =-\frac{4}{2}=-2$ and $\alpha \beta =\frac{-1}{2}=-\frac{1}{2}$.

a) We use ${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta$: $${\alpha }^2+{\beta }^2=(-2{)}^2-2\left(-\frac{1}{2}\right)=4+1=5$$