Exam practice

Statement: when a polynomial f(x) is divided by x − a, the remainder is f(a).

Proof: let the quotient be q(x) and the remainder R. By the division algorithm,

$$f(x) = (x - a)\,q(x) + R \qquad \dots \text{(i)}$$
Putting x = a in (i):
$$f(a) = (a - a)\,q(a) + R = 0 \times q(a) + R = R$$
So R = f(a). Hence proved.

$f(2) = 8 - 4 + 1 = \mathbf{5}$.

x + 1 = 0 when x = −1: $f(-1) = -1 + 5 + 6 - 16 = \mathbf{-6}$.

3x − 4 = 0 when $x = \frac{4}{3}$:
$$\begin{aligned} f\left(\tfrac{4}{3}\right) &= 6 \times \tfrac{64}{27} + 4 \times \tfrac{16}{9} + 3 \times \tfrac{4}{3} + 4 \\ &= \tfrac{128}{9} + \tfrac{64}{9} + 4 + 4 \\ &= \tfrac{192}{9} + 8 = \tfrac{64}{3} + \tfrac{24}{3} = \tfrac{88}{3} \end{aligned}$$
The remainder is $\mathbf{\frac{88}{3}}$.

2x + 6 = 0 when x = −3:
$$\begin{aligned} f(-3) &= 8(81) + 5(-27) - 2(9) + 7(-3) - 1 \\ &= 648 - 135 - 18 - 21 - 1 = 473 \end{aligned}$$
The remainder is 473.

f(−2) = 16: 2(−8) + 3(4) − k(−2) + 4 = −16 + 12 + 2k + 4 = 2k.
So 2k = 16 and k = 8.

2x − 1 = 0 when $x = \frac{1}{2}$, so $f\left(\frac{1}{2}\right) = -2$:
$$\begin{aligned} 8 \times \tfrac{1}{8} + 4 \times \tfrac{1}{4} + \tfrac{k - 3}{2} - 7 &= -2 \\ 1 + 1 + \tfrac{k - 3}{2} - 7 &= -2 \\ \tfrac{k - 3}{2} &= 3 \\ k &= 9 \end{aligned}$$
So k = 9.

f(−3) = 10:
$$\begin{aligned} -27 - 9(p - 2) + 3p + 28 &= 10 \\ -27 - 9p + 18 + 3p + 28 &= 10 \\ 19 - 6p &= 10 \\ p &= \tfrac{3}{2} \end{aligned}$$
So $\mathbf{p = \frac{3}{2}}$.

f(−2) = −16 − 4(k − 2) + 14 + 14 = 20 − 4k.
g(−2) = −24 + 32 + 2k + 6 = 14 + 2k.
Equal: 20 − 4k = 14 + 2k, so 6k = 6 and k = 1. Check: both remainders are 16. ✓

Put $x = \frac{1}{2}$:
$f\left(\frac{1}{2}\right) = \frac{1}{4} + \frac{h}{4} - 5 + 8 = \frac{h}{4} + \frac{13}{4}$ and $g\left(\frac{1}{2}\right) = \frac{3}{4} - \frac{h}{4} + 6 = \frac{27}{4} - \frac{h}{4}$.
Equal: h + 13 = 27 − h, so 2h = 14 and h = 7.

Two remainders give two equations in m and n; solve them together.

f(−2) = −8 + 4m + 2n + 6 = 24, so 4m + 2n = 26, that is 2m + n = 13 … (i)
f(3) = 27 + 9m − 3n + 6 = 54, so 9m − 3n = 21, that is 3m − n = 7 … (ii)
Adding (i) and (ii): 5m = 20, so m = 4, and from (i) n = 13 − 8 = 5.
m = 4 and n = 5.

Statement: if f(a) = 0, then x − a is a factor of f(x).

Proof: dividing f(x) by x − a gives a quotient q(x) and a remainder R, with f(x) = (x − a) q(x) + R … (i). By the remainder theorem R = f(a) = 0, so (i) becomes f(x) = (x − a) q(x). So x − a divides f(x) exactly: it is a factor. Hence proved.

x − a is a factor, so f(x) = (x − a) q(x) for some polynomial q(x) … (i). Putting x = a in (i): f(a) = (a − a) q(a) = 0 × q(a) = 0. Hence proved.

f(3) = 2(9) − 11(3) + 15 = 18 − 33 + 15 = 0. So x − 3 is a factor.

2x − 1 = 0 when $x = \frac{1}{2}$:
$$f\left(\tfrac{1}{2}\right) = 2 \times \tfrac{1}{8} - 7 \times \tfrac{1}{4} + \tfrac{13}{2} - 5 = \tfrac{1}{4} - \tfrac{7}{4} + \tfrac{26}{4} - \tfrac{20}{4} = 0$$
So 2x − 1 is a factor.

2x + 1 = 0 when $x = -\frac{1}{2}$:
$$f\left(-\tfrac{1}{2}\right) = 2\left(-\tfrac{1}{8}\right) - 5 \times \tfrac{1}{4} - \tfrac{7}{2} + 5 = -\tfrac{1}{4} - \tfrac{5}{4} - \tfrac{14}{4} + \tfrac{20}{4} = 0$$
So 2x + 1 is a factor.

f(−2) = 16 − 2k + 8 = 24 − 2k = 0, so k = 12.

f(−1) = −2 − k + 8 + 5 = 11 − k = 0, so k = 11.

Let k be added. Then f(−1) + k = 0: −1 + 6 − 11 + 1 + k = −5 + k = 0, so 5 must be added.

f(3) + k = 0: 27 + 9 − 51 + 7 + k = −8 + k = 0, so 8 must be added.

Let k be subtracted. Then f(−4) − k = 0: −64 + 144 − 104 + 30 − k = 6 − k = 0, so 6 must be subtracted.

For example f(2) = 8 − 24 + 22 − 6 = 0 and f(−1) = −1 − 6 − 11 − 6 = −24.
So 1, 2 and 3 are roots, and −1, −2, −3 are not. By the factor theorem, $x^3 - 6x^2 + 11x - 6 = \mathbf{(x - 1)(x - 2)(x - 3)}$.

Each zero gives a factor, so
$$\begin{aligned} f(x) &= (x - 2)(x - 4)(x - 6) \\ &= (x^2 - 6x + 8)(x - 6) \\ &= x^3 - 6x^2 - 6x^2 + 36x + 8x - 48 \\ &= x^3 - 12x^2 + 44x - 48 \end{aligned}$$
$f(x) = x^3 - 12x^2 + 44x - 48$. (Any multiple of it, such as $2x^3 - 24x^2 + 88x - 96$, has the same zeros; this is the simplest.) Check: f(2) = 8 − 48 + 88 − 48 = 0. ✓

f(−1) = −1 + 6 − 11 + 6 = 0, so x + 1 is a factor. Synthetic division: −1 | 1, 6, 11, 6 gives 1, 5, 6, 0.
$x^2 + 5x + 6 = (x + 2)(x + 3)$.
(x + 1)(x + 2)(x + 3)

f(2) = 8 − 12 − 8 + 12 = 0, so x − 2 is a factor. Synthetic division: 2 | 1, −3, −4, 12 gives 1, −1, −6, 0.
$x^2 - x - 6 = (x - 3)(x + 2)$.
(x − 2)(x + 2)(x − 3). (Or by grouping: $x^2(x - 3) - 4(x - 3) = (x - 3)(x^2 - 4)$.)

f(1) = 1 − 8 + 19 − 12 = 0, so x − 1 is a factor. Synthetic division: 1 | 1, −8, 19, −12 gives 1, −7, 12, 0.
$x^2 - 7x + 12 = (x - 3)(x - 4)$.
(x − 1)(x − 3)(x − 4). Check with x = 0: (−1)(−3)(−4) = −12. ✓

Multiply out: $x^3 - 11x^2 + 48x + 2x^2 - 22x + 96 - 120 = x^3 - 9x^2 + 26x - 24$.
f(2) = 8 − 36 + 52 − 24 = 0. Synthetic division: 2 | 1, −9, 26, −24 gives 1, −7, 12, 0.
$x^2 - 7x + 12 = (x - 3)(x - 4)$.
(x − 2)(x − 3)(x − 4)

Multiply out: $3x^3 - 10x^2 + 2x - 9x^2 + 30x - 6 - 10 = 3x^3 - 19x^2 + 32x - 16$.
f(1) = 3 − 19 + 32 − 16 = 0. Synthetic division: 1 | 3, −19, 32, −16 gives 3, −16, 16, 0.
$3x^2 - 16x + 16 = 3x^2 - 12x - 4x + 16 = 3x(x - 4) - 4(x - 4) = (x - 4)(3x - 4)$.
(x − 1)(x − 4)(3x − 4)

As in Example 2: f(2) = 0, the quotient is $x^2 - x - 12 = (x - 4)(x + 3)$, so (x − 2)(x − 4)(x + 3) = 0.
x = 2, −3, 4

f(2) = 16 + 12 − 22 − 6 = 0. Synthetic division: 2 | 2, 3, −11, −6 gives 2, 7, 3, 0.
$2x^2 + 7x + 3 = 2x^2 + 6x + x + 3 = (x + 3)(2x + 1)$.
(x − 2)(x + 3)(2x + 1) = 0, so $x = 2, -3, -\frac{1}{2}$.

Bring every term to one side, in descending order: $x^3 - 9x^2 + 26x - 24 = 0$.
f(2) = 8 − 36 + 52 − 24 = 0; the quotient is $x^2 - 7x + 12 = (x - 3)(x - 4)$.
x = 2, 3, 4

$x^3 - 4x^2 - 17x + 60 = 0$. f(1) = 40, f(2) = 8 − 16 − 34 + 60 = 18, f(3) = 27 − 36 − 51 + 60 = 0.
Synthetic division: 3 | 1, −4, −17, 60 gives 1, −1, −20, 0, and $x^2 - x - 20 = (x - 5)(x + 4)$.
(x − 3)(x − 5)(x + 4) = 0, so x = 3, −4, 5.

Multiply out: $x^3 - x^2 - 14x + 4x^2 - 4x - 56 + 16 = x^3 + 3x^2 - 18x - 40 = 0$.
f(−2) = −8 + 12 + 36 − 40 = 0. Synthetic division: −2 | 1, 3, −18, −40 gives 1, 1, −20, 0, and $x^2 + x - 20 = (x + 5)(x - 4)$.
x = −2, 4, −5

Multiply out and bring 10 across: $3x^3 - 19x^2 + 32x - 16 = 0$, the polynomial of 3(e).
So (x − 1)(x − 4)(3x − 4) = 0, and $x = 1, 4, \frac{4}{3}$.