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A quadratic polynomial, whose zeroes are –3 and 4, is

(A) $x^2 - x + 12$

(B) $x^2 + x + 12$

(C) $\frac{x^2}{2} - \frac{x}{2} -6$

(D) $2x^2 + 2x -24$

Answers (1)

Answer. [C]

Solution. Polynomial : It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s) and a quadratic polynomial is polynomial of degree 2.

(A) $p(x) = x^2 - x + 12$

put x = –3 put x = 4

$p(-3) = (-3)^2 - (-3) + 12$ $p(4) = (4)^2 - 4 + 12$

$= 9 + 3 + 12 = 24 \neq 0$ $= 16 - 4 + 12 = 24 \neq 0$

(B) $p(x) = x^2 + x + 12$

put x = –3 put x = 4

$p(-3) = (-3)^2 - 3 + 12$ $p(4) = (4)^2 + 4 + 12$

$= 9 + 9 = 18 \neq 0$ = 16 + 16 = 32 $\neq$ 0

(C) $p(x)=\frac{x^2}{2} - \frac{x}{2} -6$

put x = –3 put x = 4

$p(-3)=\frac{(-3)^2}{2} - \frac{-3}{2} -6$ $p(4)=\frac{(4)^2}{2} - \frac{4}{2} -6$

$=\frac{9}{2}+\frac{3}{2}-6$ $=\frac{16}{2}-2-6$

$=\frac{9+3-12}{2}=0$ $= 8 - 8 = 0$

(D) $p(x) = 2x^2 + 2x - 24$

put x = –3 put x = 4

$p(-3) = 2(-3)^2 + 2(-3) - 24$ $p(4) = 2(4)^2 + 2(4) - 24$

$= 18 -6 - 24$ $= 32 + 8 - 24$

$= -12 \neq 0$ $= 16 \neq 0$

If –3, 4 is zeros of a polynomial p(x) then p(–3) = p(4) = 0

Here only (C) option satisfy p(–3) = p(4) = 0

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