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Answer 1

Question :

If two zeros of a polynomial x³+5x²+7x+3 are -1 and -3 ,then find third zero.

Solution:

Let p(x)=$x{}^{3}+5x{}^{2}+7x+3$

Given zeroes : -1 and -3

Since, -1 and -3 are the zeroes of polynomialp(x)=x³+5x²+7x+3

Therefore,(x+3) and (x+1) are the zeros of the given polynomial.

Product of zeroes :

(x+3)(x+1)

=x²+3x+x+3

=x²+4x+3

Now divide x³+5x²+7x+3 by x²+4x+3

$\begin{array}{r|cccc} & x & + & 1 &\\ \cline{2-5} x^2 + 4x + 3 & x^3 & + 5x^2 & +7x & +3 \\ & x^3 & +4x^2 & +3x & \\ \cline{2-5} & & x^2 & + 4x & + 3\\ & & x^2 & + 4x & + 3 \\ \cline{2-5} & & & 0 &\end{array}$

since x+1 is Quotient.

Hence , the third zero of p(x) =x³+5x²+7x+3 is -1

Answer 2

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