Question

If both x-2 and $x-\dfrac{1}{2}$ are factors of $g\left( x \right)=p{{x}^{2}}+5x+r$, then
[a] p = r
[b] p+r =0
[c] 2p+r = 0
[d] p+2r =0

Answer 1

Hint: Recall factor theorem. According to factor theorem x-a is a factor of p(x) if p(a) = 0. Use the factor theorem and hence form two linear equations in p and r. Hence find the values of p and r such that x-2 and $x-\dfrac{1}{2}$ are the factors of g(x). Verify which of the above options are valid.

Complete step-by-step answer:
We have $g\left( x \right)=p{{x}^{2}}+5x+r$
Since x-2 is a factor of g(x), we have
$g\left( x \right)=\left( x-2 \right)h\left( x \right)$, where h(x) is some polynomial in x.
Put x = 2, we get
$g\left( 2 \right)=\left( 2-2 \right)h\left( 2 \right)=0\times h\left( 2 \right)=0$
Hence x = 2 is a zero of the polynomial g(x).
Now, we have $g\left( x \right)=p{{x}^{2}}+5x+r$
Substituting x=2 in the expression of g(x), we get
$g\left( 2 \right)=p{{\left( 2 \right)}^{2}}+5\left( 2 \right)+r=4p+r+10$
Hence, g(2) = 4p+r+10
But g(2) = 0
Hence, we have
\[4p+r+10=0~~~~\left( i \right)\]
Similarly, since $x-\dfrac{1}{2}$ is a factor of g(x), we have
$g\left( \dfrac{1}{2} \right)=0$
We have $g\left( x \right)=p{{x}^{2}}+5x+r$
Substituting $x=\dfrac{1}{2}$ in the expression of g(x), we get
$g\left( \dfrac{1}{2} \right)=p{{\left( \dfrac{1}{2} \right)}^{2}}+5\left( \dfrac{1}{2} \right)+r=\dfrac{p}{4}+r+\dfrac{5}{2}$
But since $g\left( \dfrac{1}{2} \right)=0$, we have
$\dfrac{p}{4}+r+\dfrac{5}{2}=0\text{ }$
Multiplying both sides by 4, we get
$p+4r+10=0\text{ }\left( ii \right)$
Multiplying equation (ii) by 4 and subtracting equation (i) from equation (ii), we get
$4p-4p+16r-r+40-10=0\Rightarrow 15r+30=0$
Subtracting 30 from both sides of the equation, we get
$15r=-30$
Dividing by 15 on both sides of the equation, we get
$r=-2$
Substituting the value of r in equation (ii), we get
$p+4\left( -2 \right)+10=0\Rightarrow p+2=0$
Subtracting 2 from both sides of the equation, we get
$p=-2$
Hence, we have p =-2 and q =-2
Hence p = r is the only correct option.
Hence option [a] is correct.

Note: Alternative solution:
We will use the property that if $\alpha ,\beta $ are the roots of the polynomial $p\left( x \right)=a{{x}^{2}}+bx+c$, then we have
$\alpha +\beta =\dfrac{-b}{a},\alpha \beta =\dfrac{c}{a}$
We have $2+\dfrac{1}{2}=\dfrac{-5}{p}\Rightarrow \dfrac{5}{2}=\dfrac{-5}{p}$
Hence, we have
$p=-2$
Also, we have
$2\times \dfrac{1}{2}=\dfrac{r}{p}\Rightarrow p=r=-2$
Hence, we have p = r = -2.
Hence option [a] is correct.