Find the remainder when \(4{x}^{3}4{x}^{2}+x5\) is divided by \(x + 1\).
5.4 Factor theorem
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5.5 Solving cubic equations

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}24032+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 \(y1\) is a factor of \(f\left(y\right)=2{y}^{4}+3{y}^{2}5y+7\).
Determine how to approach the problem
For \(y1\) 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+35+7 \\ & = 7 \end{align*}Conclusion
Since \(f\left(1\right)\ne 0\), \(y1\) is not a factor of \(f\left(y\right)=2{y}^{4}+3{y}^{2}5y+7\).
Worked example 11: Factorising cubic polynomials
Factorise completely: \(f\left(x\right)={x}^{3}+{x}^{2}9x9\)
Find a factor by trial and error
Try
\(f\left(1\right)={\left(1\right)}^{3}+{\left(1\right)}^{2}9\left(1\right)9=1+199=16\)Therefore \(\left(x1\right)\) is not a factor.
We consider the coefficients of the given polynomial and try:
\(f\left(1\right)={\left(1\right)}^{3}+{\left(1\right)}^{2}9\left(1\right)9=1+1+99=0\)
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 \({x}^{3}+{x}^{2}9x9=\left(x+1\right)\left( \ldots \right)\)
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 \({x}^{3}+{x}^{2}9x9=\left(x+1\right)\left({x}^{2}+?x9\right)\)
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
\(f\left(1\right)={\left(1\right)}^{3}2{\left(1\right)}^{2}5\left(1\right)+6=125+6=0\)Therefore \(\left(x1\right)\) is a factor.
Factorise by inspection
\({x}^{3}2{x}^{2}5x+6=\left(x1\right)\left( \ldots \right)\)
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 \({x}^{3}2{x}^{2}5x+6=\left(x1\right)\left({x}^{2}+?x6\right)\)
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 \(f\left(x\right)=\left(x1\right)\left({x}^{2}x6\right)\).
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(x1\right)\left({x}^{2}x6\right) \\ &= (x  1)(x  3)(x + 2) \end{align*}Factorising cubic polynomials
Use the factor theorem to factorise \({x}^{3}3{x}^{2}+4\) completely.
\(f\left(x\right)=2{x}^{3}+{x}^{2}5x+2\)
Find \(f\left(1\right)\).
Factorise \(f\left(x\right)\) completely.
Use the factor theorem to determine all the factors of the following expression:
\({x}^{3}+{x}^{2}17x+15\)Complete: If \(f\left(x\right)\) is a polynomial and p is a number such that \(f\left(p\right)=0\), then \(\left(xp\right)\) is...
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