Question

Find \[p\] and \[q\] such that \[\left( x+1 \right)\] and \[\left( x-1 \right)\] are factors of the polynomial \[{{x}^{4}}+p{{x}^{3}}-3{{x}^{2}}+2x+q\]. Find the value of \[a\] and \[b\] so that the polynomial \[{{x}^{3}}-a{{x}^{2}}-13x+b\] has \[\left( x-1 \right)\] and \[\left( x+3 \right)\] as factors.

Answer 1

Hint: Equate the given factor of the polynomial to zero. Find the value of \[x\] and substitute them in the polynomial. You will get two equations with \[p\] and \[q\]. Solve them to get the value of  \[p\] and \[q\]. Similarly do the same for polynomial 2 and get the value of \[a\] and \[b\].

Complete step-by-step answer:

We have been given two polynomials and their factors such that we need to find the value of \[p\],\[q,\],\[a\] and \[b\] from it. Let us take the first expression as \[f\left( x \right)\].

\[f\left( x \right)={{x}^{4}}+p{{x}^{3}}-3{{x}^{2}}+2x+q\text{ }\to (1)\]

Now we have been given that \[\left( x+1 \right)\] and \[\left( x-1 \right)\] are factors of the given polynomial. Let us equate them to zero and get the value of \[x\].

$ x+1=0\text{ and }x-1=0 $

$ \therefore x=-1\text{ and }x=1 $

Thus we got the value of \[x=1\text{ },-1\]

Let us put \[x=1\text{ }\]in equation (1)

$ f\left( x \right)={{1}^{4}}+p{{\left( 1 \right)}^{3}}-3{{\left( 1 \right)}^{2}}+2\times \left( 1 \right)+q $

$ \text{ }=1+p-3+2+q $

$ \text{ }=p+q $

Thus we get \[f\left( x \right)=p+q=0\text{ }\to \text{(2)}\]

Let us put \[x=1\text{ }\]in equation (1)

$ f\left( x \right)={{\left( -1 \right)}^{4}}+p{{\left( -1 \right)}^{3}}-3{{\left( -1 \right)}^{2}}+2\times \left( -1 \right)+q $

$ \text{ }=1-p-3-2+q $

$ \text{ }=q-p-4=0\text{ }\to \text{(3)} $

Thus we got two equations (2) and (3). Let us solve them and find the value of  \[p\] and \[q\].

From (2), \[p=-q,\] put this value in (3)

$ q-p=4 $

$ q-\left( -q \right)=4 $

$ q+q=4 $

$ 2q=4 $

$ \therefore q=2 $

Now put  \[q=2,\]

$q-p=4 $

$ 2-\left( p \right)=4 $

$  \therefore p=-2 $

Hence we got p = -2 and q = 2

Now let us take 2nd expression as \[f\left( x \right)\]

\[f\left( x \right)={{x}^{3}}-a{{x}^{2}}-13x+b\text{ }\to \text{(4)}\]

We have been given that \[\left( x-1 \right)\] and \[\left( x+3 \right)\] are the factors of this polynomial. Let us equate them to zero and get the value of \[x\].

$ x-1=0\text{ and }x+3=0 $

$ x=1\text{ and }x=-3 $

Thus we got the value of \[x=1,-3\]

Let us put  \[x= -1\] in equation (4)

$ f\left( 1 \right)={{1}^{3}}-a{{\left( 1 \right)}^{2}}-13\times 1+b $

$ \text{ =}1-a-13+b $

$  \therefore b-a=12 \text{ }\to (5) $

Now put \[x=-3\] in (4)

$ f\left( -3 \right)={{\left( -3 \right)}^{3}}-a{{\left( -3 \right)}^{2}}-13\times \left( -3 \right)+b $

$\text{ =}-27-9a+3a+b\text{ }\to (6) $

From (5) we get \[b=12+a\], put this in equation (6)

$ -27-6a+12+a=0 $

$ -15-5a=0 $

$  -5a=15 $

$a=-3 $

Now put \[a=-3\] in (5)

\[b=12-3=9\]

Thus we got \[a=-3\] and \[b=9\].

Thus we solved the given polynomials.

\[\therefore \]  \[p=-2,q=2,a=-3\text{ and }b=9\]

Note: Be careful when you substitute the value of \[x\] into the equation of the polynomial. If there is a change in sign by mistake then the value of \[p,q,a,b\] you will be wrong. So, remember to put the sign correctly.