Question

Given that \[P\left( x \right)={{x}^{4}}+a{{x}^{3}}-{{x}^{2}}+bx-12\] has factors x-2 and x+1. How do you solve the equation P(x)=0?

Answer 1

Hint: This type of problem is based on the concept of polynomials. We know that, if x-a is a factor of a polynomial, then x=a is a root of the polynomial. Here, we get 2 and -1 to be the roots of the polynomial. Substitute x=2 and x=-1 in the equation P(x) =0 and obtain two equations with constants a and b. solve the equation and find the values of a and b. Then, we get the polynomial P(x). Now multiply the two factors x-2 and x+1 and divide the obtained quadratic polynomial by P(X). We get another quadratic polynomial. Solve the quadratic polynomial and find the values of x which is the required answer.

Complete step by step solution:
According to the question, we are asked to find the values of x when P(x)=0.
We have been given \[P\left( x \right)={{x}^{4}}+a{{x}^{3}}-{{x}^{2}}+bx-12\]. ------------(1)
And the factors of P(x) are x-2 and x+1. ----------(2)
We know that for a polynomial, if x-a is a factor of the polynomial, then a is the root.
Here, x-2 is a factor and thus we get that 2 is a root of P(x).
Now, we can write x+1 as
x+1=x-(-1)
Therefore, the root is -1.
Thus, we get P(2)=0 and P(-1)=0
Let us find P(2).
\[\Rightarrow P\left( 2 \right)={{2}^{4}}+a{{2}^{3}}-{{2}^{2}}+2b-12\]
We know that \[{{2}^{4}}=16\], \[{{2}^{3}}=8\] and \[{{2}^{2}}=4\]. On substituting these values, we get
\[P\left( 2 \right)=16+8a-4+2b-12\]
On further simplification, we get
\[P\left( 2 \right)=8a-4+2b+4\]
\[\Rightarrow P\left( 2 \right)=8a+2b\]
But we know that P(2)=0.
\[\Rightarrow 8a+2b=0\]
Let us take 2 common from both the terms of the LHS.
\[\Rightarrow 2\left( 4a+b \right)=0\]
Divide the whole equation by 2.
\[\Rightarrow \dfrac{2\left( 4a+b \right)}{2}=\dfrac{0}{2}\]
On cancelling 2 from the numerator and denominator, we get
\[4a+b=\dfrac{0}{2}\]
On further simplification, we get
\[4a+b=0\]
Subtract 4a from both the sides of the equation.
\[\Rightarrow 4a+b-4a=0-4a\]
On further simplification, we get
b=-4a ----------(3)
Now, let us find P(-1).
\[\Rightarrow P\left( -1 \right)={{\left( -1 \right)}^{4}}+a{{\left( -1 \right)}^{3}}-{{\left( -1 \right)}^{2}}-b-12\]
We know that \[{{\left( -1 \right)}^{4}}=1\], \[{{\left( -1 \right)}^{3}}=-1\] and \[{{\left( -1 \right)}^{2}}=1\]. On substituting these values, we get
\[P\left( -1 \right)=1-a-1-b-12\]
On further simplification, we get
\[P\left( -1 \right)=-a-b-12\]
\[\Rightarrow P\left( -1 \right)=-a-b-12\]
But we know that P(-1)=0.
\[\Rightarrow -a-b-12=0\]
Let us take -1 common from both the terms of the LHS.
\[\Rightarrow -1\left( a+b+12 \right)=0\]
Divide the whole equation by -1.
\[\Rightarrow \dfrac{-1\left( a+b+12 \right)}{-1}=\dfrac{0}{-1}\]
On cancelling -1 from the numerator and denominator, we get
\[a+b+12=\dfrac{0}{-1}\]
On further simplification, we get
\[a+b+12=0\]
But from (3), we know that b=-4a.
On substituting the value of b in a+b+10=0, we get
a-4a+12=0
On further simplification, we get
\[-3a+12=0\]
Subtract 12 from both the sides of the equation.
\[\Rightarrow -3a+12-12=0-12\]
On further simplification, we get
-3a=-12.
Let us divide the expression by -3.
\[\Rightarrow \dfrac{-3a}{-3}=\dfrac{-12}{-3}\]
On cancelling the common terms from the numerator and denominator of both the LHS and RHS, we get
\[a=4\]
We know that b=-4a.
\[\Rightarrow b=-4\times 4\]
Therefore, we get b=-16.
Thus, the polynomial is
\[P\left( x \right)={{x}^{4}}+4{{x}^{3}}-{{x}^{2}}-16x-12\]
Now, we have to solve P(x)=0.
\[\Rightarrow {{x}^{4}}+4{{x}^{3}}-{{x}^{2}}-16x-12=0\]
Let us multiply both the factors which is \[\left( x-2 \right)\left( x+1 \right)\].
Using distributive property which is (a + b) (c + d) =ac + ad + bc + bd, we get
\[\left( x-2 \right)\left( x+1 \right)={{x}^{2}}+x-2x-2\]
On further simplification, we get
\[\left( x-2 \right)\left( x+1 \right)={{x}^{2}}-x-2\]
Now, we have to divide \[{{x}^{4}}+4{{x}^{3}}-{{x}^{2}}-16x-12\] by \[{{x}^{2}}-x-2\].
We get
\[{{x}^{2}}-x-2\overset{{{x}^{2}}+5x+6}{\overline{\left){\begin{align}
  & {{x}^{4}}+4{{x}^{3}}-{{x}^{2}}-16x-12 \\
 & \dfrac{{{x}^{4}}-{{x}^{3}}-2{{x}^{2}}}{\begin{align}
  & 5{{x}^{3}}+{{x}^{2}}-16x \\
 & \dfrac{5{{x}^{3}}-5{{x}^{2}}-10x}{\begin{align}
  & 6{{x}^{2}}-6x-12 \\
 & \dfrac{6{{x}^{2}}-6x-12}{0} \\
\end{align}} \\
\end{align}} \\
\end{align}}\right.}}\]
Therefore, we have to factorise \[{{x}^{2}}+5x+6\] to find the other two values of x.
We know that 3+2=5 and 3 multiplied by 2 is equal to 6.
Therefore, we get
\[\Rightarrow {{x}^{2}}+5x+6={{x}^{2}}+3x+2x+6\]
Take x common from the first two terms and 2 common from the last two terms.
\[\Rightarrow {{x}^{2}}+5x+6=x\left( x+3 \right)+2\left( x+3 \right)\]
Take x+3 common from the two terms.
\[\Rightarrow {{x}^{2}}+5x+6=\left( x+3 \right)\left( x+2 \right)\]
Therefore, the factors are x+3 and x+2.
We can write the factors as x-(-3) and x-(-2).
We know that if x-a is a factor of a polynomial, then a is a root of the polynomial.
Here, we get -3 and -2 as the roots of the polynomial.
Therefore, the roots of P(x)=0 are -3, -2, -1 and 2.
Hence, the values of x for P(x)=0 where \[P\left( x \right)={{x}^{4}}+a{{x}^{3}}-{{x}^{2}}+bx-12\] are -3, -2, -1 and 2.

Note: We should first find the entire factor and then find the values of x. Roots of a polynomial is the value of x for which P(x) =0. Do the long division carefully. If the reminder is not equal to 0, then you have made a calculation mistake. Avoid calculation mistakes based on sign conventions.