Question

find the zeroes of quadratic polynomial x2-8x 12 and verify the relationship between its zeroes and co-efficients

Answer 1

$Let \: polynomial \:p(x) = x^{2} - 8x + 12$

$x^{2} - 8x + 12 \\= x^{2} - 6x - 2x + 12 \\= x(x-6)-2(x-6)\\= (x-6)(x-2)$

$p(x) = 0 \:when \: x-6 = 0 \:Or \:x-2 = 0$

$i.e ., \:when \: x = 6 \:or \: x = 2$

$So, Zeroes \:of \: x^{2} - 8x + 12 \:are \: 2 \:and \:6$

Verification:

$Sum \:of \:the \: zeroes = 2 + 6 \\= 8 \\= \frac{8}{1} \\= \frac{-(coefficient \:of \:x )}{coefficient \:of \:x^{2} }$

$Product \:of \:the \: zeroes = 2 \times 6 \\= 12\\= \frac{12}{1} \\= \frac{Constant}{coefficient \:of \:x^{2} }$

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Answer 2

x^2-8x+12

=x^2-6x-2x+12

=x(x-6) -2(x-2)

• x=6 or x=2

sum if zeros =2+6 = 8

product of zero = 2×6=12

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