2.7 Polynomials, the factor theorem and algebraic division — Practise with help

1. $(x - 3)(x^2 + 2x + 1) = x^3 + 2x^2 + x - 3x^2 - 6x - 3 = x^3 - x^2 - 5x - 3$.
2. $x^2 - 2x - 8 = (x - 4)(x + 2)$.
3. $(-1)^2 + 3 \times (-1) - 4 = 1 - 3 - 4 = -6$.

If question 1 was hard, revisit expanding two brackets: multiply every term in the first bracket by every term in the second.

The divisor is $x - 1$, so substitute $x = 1$ into $p(x)$.

Work out $1^3 + 3 \times 1^2 - 2 \times 1 + 4$.

The remainder is $6$. Since the remainder is not zero, $x - 1$ is not a factor.

Substitute $x = 3$ and show that the value is zero.

Work out $3^3 - 2 \times 3^2 - 5 \times 3 + 6$ term by term.

Since $p(3) = 0$, the factor theorem gives that $x - 3$ is a factor.

The quotient is a quadratic. Try the form $x^2 + bx + c$ and match the terms after multiplying back.

Work out $(x + 1)(x^2 + c)$ by expanding and comparing the constant terms. The constant here is $-4$, and $1 \times c$ must give it.

Look for a quotient $x^2 - 4$, since $1 \times (-4) = -4$ gives the constant. Check by expanding:

The quotient is $x^2 - 4$ and the remainder is $0$, so $x + 1$ is a factor.

The constant term is $12$. Try small factors: $1$, $-1$, $2$, $-2$, $3$. Find one that gives zero.

$p(2) = 8 - 12 - 8 + 12 = 0$, so $x - 2$ is a factor. Divide to get a quadratic.

$p(2) = 0$, so $x - 2$ is a factor. Divide:

Factorise the quadratic:

So

A factor gives a zero remainder, so substitute $x = 1$ and set the result to zero.

$p(1) = 1 + k - 3 + 2 = k$. Set $k = 0$.

Substitute $x = 1$:

For $x - 1$ to be a factor, $p(1) = 0$, so

With $k = 0$, the polynomial is $x^3 - 3x + 2 = (x - 1)(x^2 + x - 2)$.

Divide by $x - 1$ to leave a quadratic, then factorise the quadratic.

Dividing gives $(x - 1)(x^2 - 5x + 6)$. Factorise $x^2 - 5x + 6$.

Divide by the known factor:

Factorise the quadratic:

So $p(x) = (x - 1)(x - 2)(x - 3)$, and the roots are $x = 1$, $x = 2$ and $x = 3$.