5.4 Factor theorem | Polynomials

5.4 Factor theorem (EMCGW)

If an integer \(a\) is divided by an integer \(b\), and the answer is \(q\) with the remainder \(r = 0\), then we know that \(b\) is a factor of \(a\).

\begin{align*} a &= b \times q + r \\ \text{and } r &= 0 \\ \text{then we know that } a &= b \times q \\ \text{and also that } \frac{a}{b} &= q \end{align*}

This is also true of polynomials; if a polynomial \(a(x)\) is divided by a polynomial \(b(x)\), and the answer is \(Q(x)\) with the remainder \(R(x) = 0\), then we know that \(b(x)\) is a factor of \(a(x)\).

\begin{align*} a(x) &= b(x) \cdot Q(x) + R(x) \\ \text{and } R(x) &= 0 \\ \text{then we know that } a(x) &= b(x) \cdot Q(x) \\ \text{and also that } \frac{a(x)}{b(x)} &= Q(x) \end{align*}

The factor theorem describes the relationship between the root of a polynomial and a factor of the polynomial.

The Factor theorem

If the polynomial \(p(x)\) is divided by \(cx - d\) and the remainder, given by \(p \left( \frac{d}{c} \right),\) is equal to zero, then \(cx - d\) is a factor of \(p(x)\).

Converse: if \((cx - d)\) is a factor of \(p(x)\), then \(p \left( \frac{d}{c} \right) = 0\).

Worked example 9: Factor theorem

Using the factor theorem, show that \(y+4\) is a factor of \(g\left(y\right)=5{y}^{4}+16{y}^{3}-15{y}^{2}+8y+16\).

Determine how to approach the problem

For \(y+4\) to be a factor, \(g\left(-4\right)\) must be equal to \(\text{0}\).

Calculate \(g\left(-4\right)\)

\begin{align*} g\left(y\right) & = 5{y}^{4}+16{y}^{3}-15{y}^{2}+8y+16 \\ \therefore g\left(-4\right) & = 5{\left(-4\right)}^{4}+16{\left(-4\right)}^{3}-15{\left(-4\right)}^{2}+8\left(-4\right)+16 \\ & = 5\left(256\right)+16\left(-64\right)-15\left(16\right)+8\left(-4\right)+16 \\ & = \text{1 280}-\text{1 024}-240-32+16 \\ & = 0 \end{align*}

Conclusion

Since \(g\left(-4\right)=\text{0}\), \(y+4\) is a factor of \(g\left(y\right)\).

In general, to factorise a cubic polynomial we need to do the following:

Find one factor by trial and error: consider the coefficients of the given cubic polynomial \(p(x)\) and guess a possible root (\(\frac{c}{d}\)).
Use the factor theorem to confirm that \(\frac{c}{d}\) is a root; show that \(p\left(\frac{c}{d}\right)\) = 0.
Divide \(p(x)\) by the factor (\(cx - d\)) to obtain a quadratic polynomial (remember to be careful with the signs).
Apply the standard methods of factorisation to determine the two factors of the quadratic polynomial.

Worked example 10: Factor theorem

Use the factor theorem to determine if \(y-1\) is a factor of \(f\left(y\right)=2{y}^{4}+3{y}^{2}-5y+7\).

Determine how to approach the problem

For \(y-1\) to be a factor, \(f\left(1\right)\) must be equal to \(\text{0}\).

Calculate \(f\left(1\right)\)

\begin{align*} f\left(y\right) & = 2{y}^{4}+3{y}^{2}-5y+7 \\ \therefore f\left(1\right) & = 2{\left(1\right)}^{4}+3{\left(1\right)}^{2}-5\left(1\right)+7 \\ & = 2+3-5+7 \\ & = 7 \end{align*}

Conclusion

Since \(f\left(1\right)\ne 0\), \(y-1\) is not a factor of \(f\left(y\right)=2{y}^{4}+3{y}^{2}-5y+7\).

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Worked example 11: Factorising cubic polynomials

Factorise completely: \(f\left(x\right)={x}^{3}+{x}^{2}-9x-9\)

Find a factor by trial and error

Try

\(f\left(1\right)={\left(1\right)}^{3}+{\left(1\right)}^{2}-9\left(1\right)-9=1+1-9-9=-16\)

Therefore \(\left(x-1\right)\) is not a factor.

We consider the coefficients of the given polynomial and try:

\(f\left(-1\right)={\left(-1\right)}^{3}+{\left(-1\right)}^{2}-9\left(-1\right)-9=-1+1+9-9=0\)

Therefore \(\left(x+1\right)\) is a factor, because \(f\left(-1\right)=0\).

Factorise by inspection

Now divide \(f\left(x\right)\) by \(\left(x+1\right)\) using inspection:

Write \({x}^{3}+{x}^{2}-9x-9=\left(x+1\right)\left( \ldots \right)\)

The first term in the second bracket must be \({x}^{2}\) to give \({x}^{3}\) and make the polynomial a cubic.

The last term in the second bracket must be \(-\text{9}\) because \((+1)(-9)=-9\).

So we have \({x}^{3}+{x}^{2}-9x-9=\left(x+1\right)\left({x}^{2}+?x-9\right)\)

Now, we must find the coefficient of the middle term:

\(\left(+1\right)\left({x}^{2}\right)\) gives the \({x}^{2}\) in the original polynomial. So, the coefficient of the \(x\)-term must be \(\text{0}\).

\(\therefore f\left(x\right)=\left(x+1\right)\left({x}^{2}-9\right)\).

Write the final answer

We can factorise the last bracket as a difference of two squares:

\begin{align*} f(x) &= \left(x+1\right)\left({x}^{2}-9\right) \\ &= (x + 1)(x - 3)(x + 3) \end{align*}

Worked example 12: Factorising cubic polynomials

Use the factor theorem to factorise \(f(x) = {x}^{3}-2{x}^{2}-5x+6\).

Find a factor by trial and error

Try

\(f\left(1\right)={\left(1\right)}^{3}-2{\left(1\right)}^{2}-5\left(1\right)+6=1-2-5+6=0\)

Therefore \(\left(x-1\right)\) is a factor.

Factorise by inspection

\({x}^{3}-2{x}^{2}-5x+6=\left(x-1\right)\left( \ldots \right)\)

The first term in the second bracket must be \({x}^{2}\) to give \({x}^{3}\) if we work backwards.

The last term in the second bracket must be \(-\text{6}\) because \((-1)(-6)=+6\).

So we have \({x}^{3}-2{x}^{2}-5x+6=\left(x-1\right)\left({x}^{2}+?x-6\right)\)

Now, we must find the coefficient of the middle term:

\(\left(-1\right)\left({x}^{2}\right)\) gives \(-{x}^{2}\). So, the coefficient of the \(x\)-term in the second bracket must be \(-\text{1}\) to give another \(-{x}^{2}\) so that overall we have \(-{x}^{2} -{x}^{2} = -2{x}^{2}\).

So \(f\left(x\right)=\left(x-1\right)\left({x}^{2}-x-6\right)\).

Make sure that the expression has been factorised correctly by checking that the coefficient of the \(x\)-term also works out: \((x)(-6) + (-1)(-x) = -5x\), which is correct.

Write the final answer

We can factorise the last bracket as:

\begin{align*} f(x) &= \left(x-1\right)\left({x}^{2}-x-6\right) \\ &= (x - 1)(x - 3)(x + 2) \end{align*}

Factorising cubic polynomials

Textbook Exercise 5.5

Find the remainder when \(4{x}^{3}-4{x}^{2}+x-5\) is divided by \(x + 1\).

\begin{align*} \text{Let } a(x)&= 4{x}^{3}-4{x}^{2}+x-5 \\ a(-1) &= 4(-1)^{3}-4(-1)^{2}+(-1)-5 \\ &= -4 -4 -1-5 \\ &= -\text{14} \end{align*}

Use the factor theorem to factorise \({x}^{3}-3{x}^{2}+4\) completely.

\begin{align*} \text{Let } a(x)&= {x}^{3}-3{x}^{2}+4 \\ a(-1) &= (-1)^{3}-3(-1)^{2}+4 \\ &= -1 -3 +4 \\ &= 0 \\ \therefore (x+1) & \text{ is a factor} \\ a(x) &= (x + 1)({x}^{2} -4x +4 ) \\ &= (x + 1)(x - 2)(x - 2) \\ &= (x + 1)(x - 2)^{2} \end{align*}

Find \(f\left(1\right)\).

\begin{align*} f(x)&= 2{x}^{3}+{x}^{2}-5x+2 \\ f(1)&= 2(1)^{3}+(1)^{2}-5(1)+2 \\ &= 2 + 1 -5 + 2 \\ &= \text{0} \end{align*}

Factorise \(f\left(x\right)\) completely.

\begin{align*} f(1)&= 0 \\ \therefore (x - 1) & \text{ is a factor of } f(x) \\ f(x) &= (x - 1)(2x^{2} + 3x - 2) \\ &= (x-1)(2x - 1)(x + 2) \end{align*}

Use the factor theorem to determine all the factors of the following expression:

\({x}^{3}+{x}^{2}-17x+15\)

\begin{align*} \text{Let } a(x) &= {x}^{3}+{x}^{2}-17x+15 \\ a(1) &= (1)^{3}+(1)^{2}-17(1)+15 \\ &= 1 +1 -17 +15 \\ &= 0 \\ \therefore a(x) &= (x-1)(x^{2} + 2x -15) \\ &= (x-1)(x+5)(x -3) \end{align*}

Complete: If \(f\left(x\right)\) is a polynomial and p is a number such that \(f\left(p\right)=0\), then \(\left(x-p\right)\) is...

a factor of \(f(x)\)

5.3 Remainder theorem

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5.5 Solving cubic equations