Answer
Verified444.9k+ views
Hint:
1) The factor theorem states that if a, b, c, … etc. are the roots of a polynomial equation p(x) = 0, then the p(a), p(b), p(c), … etc. are all equal to zero. It also means that if (x - a) is a factor of the polynomial p(x), then the value of p(a) must be equal to 0.
2) A polynomial of degree n has at most n roots.
Complete step by step solution:
Let’s say that $p(x) = {x^4} + {x^3} - 16{x^2} - 4x + 48 = 0$.
Since, the product if the roots is given to be 6, there are many possibilities for the roots, like 1 and 6, -1 and -6, 2 and 3, √12 and √3, 1.2 and 0.5, and even some complex numbers are also possible which can give the product as 6.
We will use the trial method to pick a number and check whether it is a root of the given polynomial or not by making use of the factor theorem.
Obviously, picking very large or very small values will not help because then ${x^4}$ will become either very large or very small as compared to the remaining part of the polynomial.
Let’s pick the numbers 2 and 3 (whose product is 6), and check whether they satisfy the factor theorem or not.
$p(2) = {2^4} + {2^3} - 16 \times {2^2} - 4 \times 2 + 48 = 16 + 8 - 64 - 8 + 48 = 0$.
$p(3) = {3^4} + {3^3} - 16 \times {3^2} - 4 \times 3 + 48 = 81 + 27 - 144 - 12 + 48 = 156 - 156 = 0$.
Therefore, both 2 and 3 are the roots of the polynomial.
The factor theorem also tells us that both (x - 2) and (x - 3) must be the factors of the given polynomial.
Since the degree of p(x) is 4, we must have another polynomial, say g(x), of degree 2 as its factor.
∴ ${x^4} + {x^3} - 16{x^2} - 4x + 48 = (x - 2)(x - 3)g(x)$
⇒ ${x^4} + {x^3} - 16{x^2} - 4x + 48 = ({x^2} - 5x +6)g(x)$
⇒ $g(x) = \dfrac{{x^4} + {x^3} - 16{x^2} - 4x + 48}{{x^2} - 5x +6}$.
Upon performing the long division, we get $g(x) = {x^2} + 6x + 8$.
Now, $g(x)={{x}^{2}}+4x+2x+8=x(x+4)+2(x+4)=(x+4)(x+2)$.
Finally, $p(x)=(x-2)(x-3)(x+2)(x+4)=0$, whose roots are 2, 3, -2 and -4 (by using the factor theorem).
Note:
Not every pair whose product is 6, will give us a root. For example, -2 is a factor of the given polynomial but -3 is not.
A related result of the factor theorem is the remainder theorem which states that the remainder is p(a) when p(x) is divided by (x - a).
1) The factor theorem states that if a, b, c, … etc. are the roots of a polynomial equation p(x) = 0, then the p(a), p(b), p(c), … etc. are all equal to zero. It also means that if (x - a) is a factor of the polynomial p(x), then the value of p(a) must be equal to 0.
2) A polynomial of degree n has at most n roots.
Complete step by step solution:
Let’s say that $p(x) = {x^4} + {x^3} - 16{x^2} - 4x + 48 = 0$.
Since, the product if the roots is given to be 6, there are many possibilities for the roots, like 1 and 6, -1 and -6, 2 and 3, √12 and √3, 1.2 and 0.5, and even some complex numbers are also possible which can give the product as 6.
We will use the trial method to pick a number and check whether it is a root of the given polynomial or not by making use of the factor theorem.
Obviously, picking very large or very small values will not help because then ${x^4}$ will become either very large or very small as compared to the remaining part of the polynomial.
Let’s pick the numbers 2 and 3 (whose product is 6), and check whether they satisfy the factor theorem or not.
$p(2) = {2^4} + {2^3} - 16 \times {2^2} - 4 \times 2 + 48 = 16 + 8 - 64 - 8 + 48 = 0$.
$p(3) = {3^4} + {3^3} - 16 \times {3^2} - 4 \times 3 + 48 = 81 + 27 - 144 - 12 + 48 = 156 - 156 = 0$.
Therefore, both 2 and 3 are the roots of the polynomial.
The factor theorem also tells us that both (x - 2) and (x - 3) must be the factors of the given polynomial.
Since the degree of p(x) is 4, we must have another polynomial, say g(x), of degree 2 as its factor.
∴ ${x^4} + {x^3} - 16{x^2} - 4x + 48 = (x - 2)(x - 3)g(x)$
⇒ ${x^4} + {x^3} - 16{x^2} - 4x + 48 = ({x^2} - 5x +6)g(x)$
⇒ $g(x) = \dfrac{{x^4} + {x^3} - 16{x^2} - 4x + 48}{{x^2} - 5x +6}$.
Upon performing the long division, we get $g(x) = {x^2} + 6x + 8$.
Now, $g(x)={{x}^{2}}+4x+2x+8=x(x+4)+2(x+4)=(x+4)(x+2)$.
Finally, $p(x)=(x-2)(x-3)(x+2)(x+4)=0$, whose roots are 2, 3, -2 and -4 (by using the factor theorem).
Note:
Not every pair whose product is 6, will give us a root. For example, -2 is a factor of the given polynomial but -3 is not.
A related result of the factor theorem is the remainder theorem which states that the remainder is p(a) when p(x) is divided by (x - a).
Recently Updated Pages
Identify the feminine gender noun from the given sentence class 10 english CBSE
Your club organized a blood donation camp in your city class 10 english CBSE
Choose the correct meaning of the idiomphrase from class 10 english CBSE
Identify the neuter gender noun from the given sentence class 10 english CBSE
Choose the word which best expresses the meaning of class 10 english CBSE
Choose the word which is closest to the opposite in class 10 english CBSE