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Solve ${x^4} + {x^3} - 16{x^2} - 4x + 48 = 0$, given that the product of two roots is 6.

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Last updated date: 22nd Feb 2024
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IVSAT 2024
Answer
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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).
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