Further applications of remainder and factor theorem

Certain problems arise concerning polynomials where the remainder and factor theorem are required to ascertain unknowns - factors, roots or coefficients. We review three problems here:

Example 1

Suppose $P\left(x\right)=5x^2-14x-3$P(x)=5x2−14x−3 and $Q\left(x\right)=2x^2-x+k$Q(x)=2x2−x+k (with $k$k unknown), both contain a common factor of the form $\left(x-a\right)$(x−a) where $a$a is an integer. Is it possible to determine $k$k?

Knowing $\left(x-a\right)$(x−a) is a factor of $P\left(x\right)$P(x), from the factor theorem we also know that $P\left(a\right)=0$P(a)=0.

Hence we can set $5a^2-14a-3=0$5a2−14a−3=0, and this factorises to $\left(a-3\right)\left(5a+1\right)=0$(a−3)(5a+1)=0. 

So we now know that, because $a$a is an integer, the common factor must be $\left(x-3\right)$(x−3).

Therefore, from the second polynomial $Q\left(x\right)$Q(x), we apply the factor theorem again. Specifically, we know that $Q\left(3\right)=0$Q(3)=0. This means that $2\left(3\right)^2-\left(3\right)+k=0$2(3)2−(3)+k=0.

After simplifying we thus know that $15+k=0$15+k=0, and so $k=-15$k=−15. 

Example 2

When the polynomials $P\left(x\right)=x^4+5x^3-mx+n$P(x)=x4+5x3−mx+n and $Q\left(x\right)=mx^2+nx-1$Q(x)=mx2+nx−1 are both divided by $D\left(x\right)=x-1$D(x)=x−1, the remainders are $7$7 and $-6$−6 respectively. Can we find $m$m and $n$n?

Using the remainder theorem we can develop two equations from knowing that $P\left(1\right)=7$P(1)=7 and $Q\left(1\right)=-6$Q(1)=−6. Thus, after simplifying:

$m-n$m−n	$=$=	$-1$−1
$m+n$m+n	$=$=	$-5$−5

These simultaneous equations are easily solved. By addition, $2m=-6$2m=−6 and so $m=-3$m=−3. By subtracting, $2n=-4$2n=−4 and so $n=-2$n=−2. 

Example 3 (difficult)

Suppose we know that for $10x^3+23x^2+5x-2=0$10x3+23x2+5x−2=0, one of the roots is four times another root. If we know that one of the roots is an integer, can we solve the equation?

If we call one of the roots $x=a$x=a and the other root $x=4a$x=4a, then we know that $\left(x-a\right)$(x−a) and $\left(x-4a\right)$(x−4a) are factors of $P\left(x\right)=10x^3+23x^2+5x-2$P(x)=10x3+23x2+5x−2. 

Thus, by the factor theorem, we have:

                               $P\left(a\right)=10a^3+23a^2+5a-2=0$P(a)=10a3+23a2+5a−2=0

                              $P\left(4a\right)=640a^3+368a^2+20a-2=0$P(4a)=640a3+368a2+20a−2=0

If we multiply the first of these equations by $64$64 and then subtract the second equation from it, we determine the quadratic equation $1104a^2+300a-126=0$1104a2+300a−126=0.

This quadratic equation can be factorised to $\left(2a+1\right)\left(92a-21\right)=0$(2a+1)(92a−21)=0 so that  $a=-\frac{1}{2}$a=−12​ and $a=\frac{1}{5}$a=15​ are two of the three real roots.

Note that for any cubic equation with real coefficients, knowing that two of the roots are real implies that the other root must also be real - this is a consequence of the conjugate root theorem.  

Based on the information in the question, the integer root must be $x=-2$x=−2, because $4\times\left(-\frac{1}{2}\right)=-2$4×(−12​)=−2, whereas $4\times\left(\frac{1}{5}\right)=\frac{4}{5}$4×(15​)=45​. 

Hence we can write that $P\left(x\right)=\left(x+2\right)\left(2x+1\right)\left(5x-1\right)$P(x)=(x+2)(2x+1)(5x−1) with the roots of $P\left(x\right)=0$P(x)=0 given by $x=-2,-\frac{1}{2},\frac{1}{5}$x=−2,−12​,15​.

 

 

 

Worked Examples

Question 1

Question 2

Question 3