Polynomials: past paper questions & worked solutions

The curve touches the x-axis at $x=-\frac32$, so this is a repeated root. It crosses the x-axis at $x=4$, so this is a simple root. Hence $$f(x)=k(2x+3)^2(x-4).$$

The curve passes through $(0,-12)$. Substituting $x=0$ gives $$-12=k(3)^2(-4)=-36k,$$ so $k=\frac13$. Therefore $$f(x)=\frac13(2x+3)^2(x-4).$$

At an intersection with the line $y=2x-12$, $$\frac13(2x+3)^2(x-4)=2x-12.$$ Multiplying by 3 and expanding gives $$(4x^2+12x+9)(x-4)=6x-36,$$ so $$4x^3-4x^2-45x=0.$$

Factor out the known root $x=0$: $$x(4x^2-4x-45)=0.$$ The two further intersections satisfy $$4x^2-4x-45=0.$$ Using the quadratic formula, $$x=\frac{4\pm\sqrt{(-4)^2-4(4)(-45)}}{8} =\frac{4\pm4\sqrt{46}}{8} =\frac{1\pm\sqrt{46}}{2}.$$

The roots of $f(x)=0$ are $x=-2$ and $x=5$, with $x=5$ repeated. For $f\left(\dfrac13x\right)=0$, set $\dfrac13x=-2$ or $5$, giving $x=-6$ or $15$.

For $f(x-3)=0$, set $x-3=-2$ or $5$, giving $x=1$ or $8$.

Since the curve cuts at $(-2,0)$ and touches at $(5,0)$, write $f(x)=k(x+2)(x-5)^2$.

The curve passes through $(0,10)$, so $10=k(2)(25)=50k$.

$k=\dfrac15$, hence $f(x)=\dfrac15(x+2)(x-5)^2$.

$\dfrac15(x+2)(x-5)^2=10(x+2)$.

The factor $x+2$ gives one solution $x=-2$.

For $x\ne-2$, divide by $x+2$: $\dfrac15(x-5)^2=10$, so $(x-5)^2=50$.

$x=5\pm\sqrt{50}=5\pm5\sqrt2$. Therefore the three x-coordinates are $-2$, $5-5\sqrt2$ and $5+5\sqrt2$.

$3x^2+6x+9=3(x^2+2x)+9=3\big((x+1)^2-1\big)+9=3(x+1)^2+6$.

The minimum occurs when $(x+1)^2=0$, so $x=-1$ and $y=6$. Therefore $P=(-1,6)$.

Since $C_2$ intersects the x-axis at $-4,-2$ and $3$, it has form $y=k(x+4)(x+2)(x-3)$.

Substitute $P=(-1,6)$: $6=k(3)(1)(-4)=-12k$, so $k=-\dfrac12$.

$(x+4)(x+2)(x-3)=(x^2+6x+8)(x-3)=x^3+3x^2-10x-24$.

Thus $y=-\dfrac12x^3-\dfrac32x^2+5x+12$, so $A=-\dfrac12$, $B=-\dfrac32$, $C=5$ and $D=12$.

Since $C_1$ cuts the x-axis at $0$ and $4$, let $f(x)=ax(x-4)$.

At $x=2$, $f(2)=-4a=-4.8=-\dfrac{24}{5}$, so $a=\dfrac65$.

$f(x)=\dfrac65x(x-4)$.

The cubic cuts the x-axis at the origin and meets the x-axis at $x=4$, so $x=4$ is a repeated root. Let $g(x)=bx(x-4)^2$.

$7.2=b(6)(2^2)=24b$, so $b=0.3=\dfrac3{10}$.

$\dfrac65x(x-4)=\dfrac3{10}x(x-4)^2$.

Ignoring the shared roots on the x-axis, divide by $x(x-4)$ to get $\dfrac65=\dfrac3{10}(x-4)$. Hence $x=8$.

$y=f(8)=\dfrac65(8)(4)=\dfrac{192}{5}$.

$f(-1)=-a+(6a+8)+a^2=a^2+5a+8$. Since $f(-1)=32$, $a^2+5a-24=0$.

$a^2+5a-24=(a-3)(a+8)=0$. Since $a$ is positive, the only possible value is $a=3$.

Using $a=3$, $f(x)=3x^3+26x^2-9x=x(3x^2+26x-9)=x(3x-1)(x+9)$.

$x=0$, $x=\dfrac13$, or $x=-9$.

Let $u=y^{1/3}$. Then $y=u^3$ and $y^{2/3}=u^2$, so $3y+26y^{2/3}-9y^{1/3}=0$ becomes $3u^3+26u^2-9u=0$.

From part (a), $u=-9,0,\dfrac13$. Therefore $y=u^3$, giving $y=-729,0,\dfrac1{27}$.

Let $u=9^z$. Then $3(9^{3z})+26(9^{2z})-9(9^z)=0$ becomes $3u^3+26u^2-9u=0$.

Since $9^z>0$, the only valid root is $u=\dfrac13$. Hence $9^z=\dfrac13$, so $(3^2)^z=3^{-1}$ and $2z=-1$. Therefore $z=-\dfrac12$.