Polynomials: Division, Remainder and Factor Theorems: past paper questions & worked solutions

Dividing by $(x-2)$ means the remainder is $f(2)$.

$f(2)=a(2)^3+3(2)^2-8(2)+2=8a+12-16+2=8a-2$.

The remainder is $3$, so $8a-2=3$.

$8a=5$, hence $a=\dfrac58$.

Since $(2x-3)$ is a factor, $x=\dfrac32$ is a root, so $f\left(\dfrac32\right)=0$.

$4\left(\dfrac32\right)^3-8\left(\dfrac32\right)^2+5\left(\dfrac32\right)+a=0$.

$13.5-18+7.5+a=0$, so $3+a=0$ and $a=-3$.

With $a=-3$, $f(x)=4x^3-8x^2+5x-3$.

$4x^3-8x^2+5x-3=(2x-3)(2x^2-x+1)$.

The discriminant of $2x^2-x+1$ is $(-1)^2-4(2)(1)=1-8=-7$.

Since the quadratic has negative discriminant, it has no real roots. Therefore $f(x)=0$ has only the one real root from $2x-3=0$, namely $x=\dfrac32$.

Dividing by $(x+1)$ means use $x=-1$. Hence $f(-1)=1-a-3-b+5=3-a-b$.

The remainder is $4$, so $3-a-b=4$. Therefore $a+b=-1$.

Dividing by $(x-2)$ means use $x=2$. Hence $f(2)=16+8a-12+2b+5=9+8a+2b$.

The remainder is $-23$, so $9+8a+2b=-23$, which gives $4a+b=-16$.

Subtract $a+b=-1$ from $4a+b=-16$ to get $3a=-15$, so $a=-5$.

Using $a+b=-1$, $-5+b=-1$, hence $b=4$.

Since $(x+2)$ is a factor of $f(x)=4x^3+6x+k$, we have $f(-2)=0$.

$4(-2)^3+6(-2)+k=0$, so $-32-12+k=0$ and $k=44$.

The remainder on division by $(x-5)$ is $f(5)=4(5)^3+6(5)+44=574$.

Write $6x^3-15x^2-21x+8=(2x+3)(Ax^2+Bx+C)+R$.

Comparing coefficients gives $A=3$, $B=-12$, and $C=\dfrac{15}{2}$.

Then $(2x+3)\left(3x^2-12x+\dfrac{15}{2}\right)=6x^3-15x^2-21x+\dfrac{45}{2}$, so $R=8-\dfrac{45}{2}=-\dfrac{29}{2}$.

Dividing by $x+3$ leaves remainder $55$, so $f(-3)=55$.

$f(-3)=2(-3)^3-(-3)^2-3A+B=-63-3A+B$. Therefore $-63-3A+B=55$, so $3A-B=-118$.

Since $2x-5$ is a factor, $x=\dfrac52$ is a root and $f\left(\dfrac52\right)=0$.

$2\left(\dfrac52\right)^3-\left(\dfrac52\right)^2+\dfrac52A+B=0$, so $25+\dfrac52A+B=0$. Multiplying by $2$ gives $5A+2B=-50$.

From $3A-B=-118$, $B=3A+118$. Substitute into $5A+2B=-50$ to get $11A=-286$, so $A=-26$ and $B=40$.

$f(x)=2x^3-x^2-26x+40$.

$2x^3-x^2-26x+40=(x-7)(2x^2+13x+65)+495$, so the quotient is $2x^2+13x+65$.

Since $(x+2)$ is a factor, $f(-2)=0$.

$A(-2)^3+6(-2)^2-4(-2)+B=0$, so $-8A+32+B=0$ and $B=8A-32$.

$\int_3^5(Ax^3+6x^2-4x+B)\,dx=176$.

$\left[\dfrac A4x^4+2x^3-2x^2+Bx\right]_3^5=176$.

This gives $136A+164+2B=176$, so $68A+B=6$.

Using $B=8A-32$ in $68A+B=6$ gives $76A=38$, so $A=\dfrac12$.

$B=8\left(\dfrac12\right)-32=-28$.

$x^2+3=13-\dfrac9{x^2}$.

Multiplying by $x^2$ gives $x^4+3x^2=13x^2-9$, so $x^4-10x^2+9=0$.

Let $u=x^2$. Then $u^2-10u+9=(u-1)(u-9)=0$, so $x^2=1$ or $x^2=9$.

Since $x>0$, the intersections have $x=1$ and $x=3$. Thus $x_P=1$ and $x_Q=3$.

Between $x=1$ and $x=3$, curve $C_2$ is above curve $C_1$, so $R=\displaystyle\int_1^3\left(13-\dfrac9{x^2}-(x^2+3)\right)dx$.

$R=\displaystyle\int_1^3(10-x^2-9x^{-2})\,dx$.

$R=\left[10x-\dfrac{x^3}{3}+\dfrac9x\right]_1^3$.

$R=\left(30-9+3\right)-\left(10-\dfrac13+9\right)=24-\dfrac{56}{3}=\dfrac{16}{3}$.

The remainder on division by $(x-2)$ is $f(2)$.

$f(2)=(2-2)(2(2)^2+5(2)+k)+21=21$.

Since $(2x-1)$ is a factor, $x=\dfrac12$ is a root, so $f\left(\dfrac12\right)=0$.

$\left(\dfrac12-2\right)\left(2\left(\dfrac12\right)^2+5\left(\dfrac12\right)+k\right)+21=0$.

$-\dfrac32(3+k)+21=0$. Multiplying by $2$ gives $-3(k+3)+42=0$, so $k=11$.

$f(x)=(x-2)(2x^2+5x+11)+21=2x^3+x^2+x-1$.

$2x^3+x^2+x-1=(2x-1)(x^2+x+1)$.

The quadratic $x^2+x+1$ has discriminant $1-4=-3$, so it has no real roots. Therefore $f(x)=0$ has one real solution.

$2\log_4(x+3)=\log_4((x+3)^2)$, so the left side is $\log_4(x(x+3)^2)$.

$\dfrac12=\log_4 2$, so the right side is $\log_4(4x+2)+\log_4 2=\log_4(8x+4)$.

Therefore $x(x+3)^2=8x+4$.

$x(x^2+6x+9)=8x+4$, so $x^3+6x^2+9x=8x+4$.

Rearranging gives $x^3+6x^2+x-4=0$.

Since $-1$ is a root, $(x+1)$ is a factor of $x^3+6x^2+x-4$.

$x^3+6x^2+x-4=(x+1)(x^2+5x-4)$.

$x^2+5x-4=0$ gives $x=\dfrac{-5\pm\sqrt{25+16}}{2}=\dfrac{-5\pm\sqrt{41}}{2}$.

The original log equation requires $x>0$. The only positive root is $x=\dfrac{-5+\sqrt{41}}{2}$.

At the $y$-axis, $x=0$. On $C$, $y=0^2-4(0)+5=5$, so $D=(0,5)$.

The line is $y=2$, so solve $x^2-4x+5=2$.

$x^2-4x+3=0$, so $(x-1)(x-3)=0$. Hence $x_E=1$ and $x_F=3$.

$R_1$ lies between the curve and the line $y=2$ from $x=0$ to $x=1$.

$\displaystyle R_1=\int_0^1\left(x^2-4x+5-2\right)dx=\int_0^1(x^2-4x+3)dx$.

$\displaystyle R_1=\left[\dfrac{x^3}{3}-2x^2+3x\right]_0^1=\dfrac13-2+3=\dfrac43$.

The line through $D=(0,5)$ and $F=(3,2)$ has gradient $-1$, so its equation is $y=5-x$.

For $0\le x\le1$, $R_2$ is between $y=5-x$ and the curve. For $1\le x\le3$, it is between $y=5-x$ and $y=2$.

$\displaystyle R_2=\int_0^1\left[(5-x)-(x^2-4x+5)\right]dx+\int_1^3\left[(5-x)-2\right]dx$.

$\displaystyle R_2=\int_0^1(-x^2+3x)dx+\int_1^3(3-x)dx=\dfrac{7}{6}+2=\dfrac{19}{6}$.

$k=\dfrac{\text{area of }R_1}{\text{area of }R_2}=\dfrac{4/3}{19/6}=\dfrac{8}{19}$.

$f(1)=3+A+B-10=A+B-7$, so $k=A+B-7$.

$f(-1)=-3+A-B-10=A-B-13$.

Since $f(-1)=-10k$, $A-B-13=-10(A+B-7)$.

Rearranging gives $11A+9B=83$.

Since $(3x-2)$ is a factor, $f\left(\dfrac23\right)=0$.

$3\left(\dfrac23\right)^3+A\left(\dfrac23\right)^2+B\left(\dfrac23\right)-10=0$. Multiplying by $9$ gives $2A+3B=41$.

Solving $11A+9B=83$ with $2A+3B=41$ gives $A=-8$ and $B=19$.

With these values, $f(x)=3x^3-8x^2+19x-10=(3x-2)(x^2-2x+5)$.

Dividing by $(2x-3)$ means use $x=\dfrac32$.

$f\left(\dfrac32\right)=\left(\dfrac94-2\right)(0)-21=-21$.

$f(3)=(9-2)(6-3)-21=7\cdot3-21=0$.

Since $f(3)=0$, $(x-3)$ is a factor of $f(x)$.

$f(x)=(x^2-2)(2x-3)-21=2x^3-3x^2-4x-15$.

Dividing by $(x-3)$ gives $2x^2+3x+5$, so $f(x)=(x-3)(2x^2+3x+5)$.

For $2x^2+3x+5$, the discriminant is $3^2-4(2)(5)=9-40=-31<0$.

The quadratic has no real roots, so $f(x)=0$ has only one real root, $x=3$.

For $(2x+3)$, use $x=-\dfrac32$. Substituting gives $f\left(-\dfrac32\right)=6\left(-\dfrac32\right)^3+17\left(-\dfrac32\right)^2+4\left(-\dfrac32\right)-12=0$.

Since $f\left(-\dfrac32\right)=0$, $(2x+3)$ is a factor of $f(x)$.

$6x^3+17x^2+4x-12=(2x+3)(3x^2+4x-4)$.

$3x^2+4x-4=(3x-2)(x+2)$, so $f(x)=(2x+3)(3x-2)(x+2)$.

Let $t=\tan\theta$. Then $(2t+3)(3t-2)(t+2)=0$, so $t=-\dfrac32$, $t=\dfrac23$, or $t=-2$.

For $\dfrac\pi2<\theta<\pi$, $\tan\theta$ is negative, so use $\tan\theta=-\dfrac32$ and $\tan\theta=-2$.

This gives $\theta=\pi-\tan^{-1}\left(\dfrac32\right)=2.16$ and $\theta=\pi-\tan^{-1}(2)=2.03$, to 3 significant figures.

When $f(x)$ is divided by $(x-3)$, the remainder is $f(3)$. Since $f(3)=(3-3)(3\cdot3^2+3+a)-35$, the remainder is $-35$.

Since $(3x-2)$ is a factor, $x=\dfrac23$ is a root and $f\left(\dfrac23\right)=0$.

$0=\left(\dfrac23-3\right)\left(3\left(\dfrac23\right)^2+\dfrac23+a\right)-35=-\dfrac73(a+2)-35$. Hence $-7(a+2)=105$, so $a=-17$.

$f(x)=(x-3)(3x^2+x-17)-35=3x^3-8x^2-20x+16$.

Dividing $3x^3-8x^2-20x+16$ by $(3x-2)$ gives $x^2-2x-8$.

$x^2-2x-8=(x+2)(x-4)$, so $f(x)=(3x-2)(x+2)(x-4)$.

Since $x+3$ is a factor, $f(-3)=0$.

$f(-3)=-27a+9b-15-3=0$, so $9b=27a+18$ and therefore $b=3a+2$.

When dividing by $2x-1$, substitute $x=\dfrac12$. The remainder is $\dfrac74$, so $f\left(\dfrac12\right)=\dfrac74$.

$\dfrac a8+\dfrac b4+\dfrac52-3=\dfrac74$. Multiplying by $8$ gives $a+2b-4=14$, so $a+2b=18$.

Substitute $b=3a+2$ into $a+2b=18$: $a+6a+4=18$, so $a=2$ and $b=8$.

$f(x)=2x^3+8x^2+5x-3$.

$2x^3+8x^2+5x-3=(x-2)(2x^2+12x+29)+55$.

The quotient is $2x^2+12x+29$ and the remainder is $55$.

Since $3x-4$ is a factor, $x=\dfrac43$ is a root and $f\left(\dfrac43\right)=0$.

$3\left(\dfrac43\right)^3+a\left(\dfrac43\right)^2-10\left(\dfrac43\right)+b=0$, so $\dfrac{64}{9}+\dfrac{16a}{9}-\dfrac{120}{9}+b=0$. Multiplying by $9$ gives $16a+9b=56$.

When dividing by $x-2$, the remainder is $f(2)$. Given that the remainder is $b$, $f(2)=b$.

$f(2)=24+4a-20+b=4+4a+b$. Thus $4+4a+b=b$, so $a=-1$.

Substitute $a=-1$ into $16a+9b=56$: $-16+9b=56$, so $b=8$.

$f(x)=3x^3-x^2-10x+8$.

$f(1)=3-1-10+8=0$, so $(x-1)$ is a factor.

Dividing by $(x-1)$ gives $3x^2+2x-8=(3x-4)(x+2)$. Hence $f(x)=(x-1)(x+2)(3x-4)$.

Since $(2x+3)$ is a factor, $x=-\dfrac32$ is a root.

$4\left(-\dfrac32\right)^3+p\left(-\dfrac32\right)^2+8\left(-\dfrac32\right)+q=0$.

$-\dfrac{27}{2}+\dfrac94p-12+q=0$, so $\dfrac94p+q=\dfrac{51}{2}$.

The remainder on division by $(x+2)$ is $f(-2)$.

$f(-2)=4(-8)+4p-16+q=4p+q-48$.

$4p+q-48=-5$, so $4p+q=43$.

Subtract $\dfrac94p+q=\dfrac{51}{2}$ from $4p+q=43$: $\dfrac74p=\dfrac{35}{2}$, so $p=10$. Then $4(10)+q=43$, so $q=3$.

With $p=10$, $f(x)=4x^3+10x^2+8x+3$, so $f'(x)=12x^2+20x+8$.

$f'(x)=4(3x+2)(x+1)$.

The quadratic derivative is negative between its roots, so $f(x)$ is decreasing for $-1<x<-\dfrac23$.

Substitute $t=4$: $3\log_2(8)-2\log_2(2)=3(3)-2(1)=7$.

$3\log_2(t+4)-2\log_2(t-2)=\log_2\left(\dfrac{(t+4)^3}{(t-2)^2}\right)$.

$7=\log_2(128)$, so $\dfrac{(t+4)^3}{(t-2)^2}=128$.

$(t+4)^3=128(t-2)^2$.

$t^3+12t^2+48t+64=128t^2-512t+512$.

Rearranging gives $t^3-116t^2+560t-448=0$.

Since $t=4$ is a root, $(t-4)$ is a factor.

$t^3-116t^2+560t-448=(t-4)(t^2-112t+112)$.

$t^2-112t+112=0$ gives $t=\dfrac{112\pm\sqrt{112^2-4(112)}}{2}=56\pm12\sqrt{21}$.

The original equation requires $t>2$. The value $56-12\sqrt{21}$ is less than $2$, so it is rejected. Hence $t=4$ or $t=56+12\sqrt{21}$.

Since $(2x+1)$ is a factor, $x=-\dfrac12$ is a root, so $f\left(-\dfrac12\right)=0$.

$4\left(-\dfrac12\right)^3+a\left(-\dfrac12\right)^2-29\left(-\dfrac12\right)+b=0$.

$-\dfrac12+\dfrac a4+\dfrac{29}{2}+b=0$. Multiplying by $4$ gives $a+4b+56=0$, so $a+4b=-56$.

Dividing by $(x-2)$ gives remainder $f(2)$.

$f(2)=4(8)+4a-58+b=4a+b-26$.

The remainder is $-25$, so $4a+b-26=-25$, hence $4a+b=1$.

Solve $a+4b=-56$ and $4a+b=1$. Multiplying the second equation by $4$ gives $16a+4b=4$.

Subtracting $a+4b=-56$ from $16a+4b=4$ gives $15a=60$, so $a=4$. Then $16+b=1$, so $b=-15$.

Therefore $f(x)=4x^3+4x^2-29x-15$.

Using the known factors $(2x+1)$ and $(x-2)$, division gives $f(x)=(2x+1)(x-2)(2x+15)$.

Since $(x-3)$ is a factor, $f(3)=0$.

$27+9(p+3)-3+q=0$, so $51+9p+q=0$.

Therefore $9p+q=-51$.

Dividing by $(x+p)$ means substituting $x=-p$. The remainder is $f(-p)$.

$f(-p)=-p^3+(p+3)p^2+p+q=3p^2+p+q$.

The remainder is $9$, so $3p^2+p+q=9$, hence $3p^2+p+q-9=0$.

From $9p+q=-51$, $q=-51-9p$. Substitute into $3p^2+p+q-9=0$.

$3p^2+p-51-9p-9=0$, so $3p^2-8p-60=0$. This gives $p=6$ or $p=-\dfrac{10}{3}$.

Since $p>0$, $p=6$. Then $q=-51-9(6)=-105$.

$f(x)=x^3+9x^2-x-105$.

$x^3+9x^2-x-105=(x-3)(x^2+12x+35)$, so $g(x)=x^2+12x+35$.

$3\log_3(2x-1)=\log_3(2x-1)^3$.

Since $2=\log_3 9$, the right side is $\log_3 9+\log_3(14x-25)=\log_3\{9(14x-25)\}$.

So $(2x-1)^3=9(14x-25)$.

$8x^3-12x^2+6x-1=126x-225$, which simplifies to $2x^3-3x^2-30x+56=0$.

Substituting $x=-4$ gives $2(-64)-3(16)-30(-4)+56=0$, so $-4$ is a root.

Dividing by $(x+4)$ gives $2x^2-11x+14=(2x-7)(x-2)$.

The possible roots are $-4$, $2$, and $\dfrac72$. The original logarithms require $2x-1>0$ and $14x-25>0$, so $x>\dfrac{25}{14}$.

Therefore the solutions are $x=2$ and $x=\dfrac72$.

$4x^3+13x^2-10x+8=(x-2)(4x^2+21x+32)+72$.

Hence $Q(x)=4x^2+21x+32$ and $R=72$.

$f(-4)=4(-4)^3+13(-4)^2-10(-4)+8=-256+208+40+8=0$.

Since $f(-4)=0$, $(x+4)$ is a factor of $f(x)$.

Dividing by $(x+4)$ gives $f(x)=(x+4)(4x^2-3x+2)$.

The discriminant of $4x^2-3x+2$ is $(-3)^2-4(4)(2)=9-32=-23<0$, so this quadratic has no real roots.

Therefore $f(x)=0$ has only one real solution, $x=-4$.

$f'(x)=12x^2+26x-10=2(3x-1)(2x+5)$.

$f(x)$ is decreasing when $f'(x)<0$, which is between the roots. Hence $-\dfrac52<x<\dfrac13$.

The circle has centre $(3,5)$ and radius $r$, so $(x-3)^2+(y-5)^2=r^2$.

On the line, $y=2x+k$. Substitution gives $(x-3)^2+(2x+k-5)^2=r^2$.

$x^2-6x+9+4x^2+4kx-20x+k^2-10k+25-r^2=0$.

Collecting terms gives $5x^2+(4k-26)x+k^2-10k+34-r^2=0$.

For a tangent, the quadratic has one repeated root, so its discriminant is zero.

$(4k-26)^2-20(k^2-10k+34-r^2)=0$.

This reduces to $20r^2-4k^2-8k-4=0$, hence $5r^2=(k+1)^2$. Therefore $p=1$.

For the repeated root, $x_B=-\dfrac{4k-26}{2\cdot5}=\dfrac{13-2k}{5}$.

The point $A$ is $(0,k)$ and $B$ lies on a line of gradient $2$, so $AB=\sqrt{x_B^2+(2x_B)^2}=\sqrt5\,x_B$.

From $5r^2=(k+1)^2$, $r=\dfrac{k+1}{\sqrt5}$ in the shown configuration.

$\sqrt5\left(\dfrac{13-2k}{5}\right)=2\left(\dfrac{k+1}{\sqrt5}\right)$.

Multiplying by $\sqrt5$ gives $13-2k=2k+2$, so $k=\dfrac{11}{4}$.