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# How do you find the zeros $p(x) = {x^4} - 2{x^3} - 7{x^2} + 8x + 12$?

Last updated date: 22nd Jul 2024
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Hint: A polynomial is an articulation that can be worked from constants and images called factors or indeterminates by methods for expansion, augmentation and exponentiation to a non-negative number force.

Let
\eqalign{ & f(x) = {x^4} - 2{x^3} - 7{x^2} + 8x + 12 \cr & \Rightarrow f( - 1) = {( - 1)^4} - 2{( - 1)^3} - 7{( - 1)^2} + 8( - 1) + 12 = 0 \cr}
So, is a factor.
Using synthetic division
\eqalign{ & - 1|1 - 2 - 7812 \cr & |0 - 134 - 12 \cr & |1. - 3 - 4120* \cr}
Where $0*$ is a zero only but it indicates the remainder of the division.
So, the polynomial is written as
$\Rightarrow {x^4} - 2{x^3} - 7{x^2} + 8x + 12 = (x + 1)({x^3} - 3{x^2} - 4x + 12)$
Now, consider
$\Rightarrow g(x) = {x^3} - 3{x^2} - 4x + 12$
Similarly if we test as above
\eqalign{ & \Rightarrow g( - 2) = {( - 2)^3} - 3{( - 2)^2} - 4( - 2) + 12 \cr & \Rightarrow - 8 - 12 + 8 + 12 = 0 \cr}
Now again computing synthetic division
\eqalign{ & - 2|1 - 3 - 4.12 \cr & |0 - 210 - 12 \cr & |1 - 560 \cr}
So, the
\eqalign{ & \Rightarrow g(x) = (x + 2)({x^2} - 5x + 6) \cr & \Rightarrow {x^2} - 5x + 6 = {x^2} - 2x - 3x + 6 \cr & \Rightarrow x(x - 2) - 3(x - 2) \cr & \Rightarrow (x - 2)(x - 3) \cr & \Rightarrow g(x) = (x + 2)(x - 2)(x - 3) \cr & \Rightarrow f(x) = {x^4} - 2{x^3} - 7{x^2} + 8x + 12 \cr & \Rightarrow f(x) = (x + 1)g(x) \cr & \Rightarrow f(x) = (x + 1)(x + 2)(x - 2)(x - 3) \cr}
$\therefore$So, the roots are $x = 1,2, - 2,3$

Note: Polynomials show up in numerous zones of arithmetic and science .The x happening in a polynomial is usually called a variable or a vague. At the point when the polynomial is considered as an articulation, x is a fixed image which doesn't have any worth.