3B, 3C - Factorising and solving cubics

Worked example 1: a supplied monic factor

Question. Given that $x-2$ is a factor of $f(x)=x^3+2x^2-5x-6$, factorise $f(x)$ completely.

Plan. Divide by $x-2$, then factorise the quadratic quotient.

Therefore $f(x)=(x-2)(x^2+4x+3)$. The quadratic needs two numbers with sum $4$ and product $3$: $1$ and $3$.

Answer. $f(x)=(x-2)(x+1)(x+3)$.

Check. Expanding the last two factors gives $x^2+4x+3$. Multiplying by $x-2$ gives $x^3+2x^2-5x-6$, including the original leading coefficient and constant.

Worked example 2: a supplied non-monic factor

Question. Given that $2x-1$ is a factor of $f(x)=2x^3+3x^2-8x+3$, factorise $f(x)$ completely.

Plan. Divide by the whole factor $2x-1$, then factorise the quotient.

The quotient is $x^2+2x-3$. Since $3+(-1)=2$ and $3(-1)=-3$,

Answer. $(2x-1)(x+3)(x-1)$.

Check. Expanding the product gives $2x^3+3x^2-8x+3$. The leading term is $2x^3$, as required.

Watch out

Replacing $2x-1$ by $x-1/2$ in a factorisation loses a factor of $2$. If you use the latter form, write $2(x-1/2)$. Check the leading coefficient as well as the roots.

Worked example 3: finding an integer root

Question. Find a rational root of $f(x)=x^3-4x^2-x+4$, then express $f(x)$ as a linear factor times a quadratic.

Plan. Try $1$, a divisor of the constant $4$, and divide if the result is zero.

Therefore $x-1$ is a factor. Long division gives

Answer. A rational root is $1$, and $f(x)=(x-1)(x^2-3x-4)$.

Check. Expanding this product gives $x^3-4x^2-x+4$, independently confirming the factor and quotient.

Worked example 4: finding a fractional root

Question. Find a rational root of $f(x)=2x^3+x^2-7x+3$, then express $f(x)$ as a linear factor times a quadratic.

Plan. Test candidates whose numerator divides $3$ and whose denominator divides $2$, then divide by the corresponding linear factor.

The candidates in lowest terms are $\pm1$, $\pm3$, $\pm1/2$ and $\pm3/2$. First,

Trying the small positive fraction gives

Hence $2x-1$ is a factor. Divide:

Answer. A rational root is $1/2$, and $f(x)=(2x-1)(x^2+x-3)$.

Check. Expanding gives $2x^3+x^2-7x+3$. At $x=2$, both the original polynomial and the product give $9$, a further arithmetic check.

Problem-solving

Test cheap candidates first, record the substitution that gives zero, then divide once. A rounded calculator value close to zero is not a proof of an exact factor.

You already know

For real numbers, a product is zero if and only if at least one factor is zero. A quadratic can be solved by factorisation, completing the square or the quadratic formula.

Worked example 5: irrational roots from the quadratic

Question. Solve $x^3-5x-2=0$, given that $x+2$ is a factor. Give all real solutions exactly.

Plan. Divide by $x+2$, then solve both the linear and quadratic factors.

Division gives

The linear factor gives $x=-2$. Complete the square in the quadratic:

Answer. $x=-2$, $x=1+\sqrt2$ or $x=1-\sqrt2$.

Check. At $x=-2$, the original cubic is $-8+10-2=0$. For either $x=1\pm\sqrt2$, the relation $x^2=2x+1$ gives $x^3=5x+2$, so the original cubic is again zero. This checks both irrational roots without rounding.

Worked example 6: a repeated root

Question. Solve $x^3-3x+2=0$, given that $x+2$ is a factor. State any repeated root.

Plan. Divide, factorise the quotient and apply the zero-product rule.

Thus $x+2=0$ or $x-1=0$.

Answer. The distinct real solutions are $x=-2$ and $x=1$. The root $1$ is repeated, because its factor occurs twice.

Check. Direct substitution gives $(-2)^3-3(-2)+2=0$ and $1^3-3(1)+2=0$. The quadratic discriminant is $(-2)^2-4(1)(1)=0$, confirming its repeated root.

Key idea

One linear factor unlocks the cubic, but the remaining quadratic determines whether the other roots are distinct real values, a repeated value or non-real. State every real value once and identify repetition when asked.