Question

find the zeros of the polynomial x^2+7x-8 and verify the relationship between the zeros and the coefficient

Answer 1

HERE'S YOUR ANSWER
p(x)=x^2+7x-8
{by middle term splitting}
=x^2-x+8x-8
x(x-1)+8(x-1)
=(x+8)(x-1)=p(x)
PUT p(x)=0
x+8=0 x-1=0
x=-8. x=1
therefore two zeroes of p(x) are -8 and 1

VERIFICATION
$\alpha + \beta = \frac{ - b}{a } \\ - 8 + 1 = \frac{ - (7)}{1} \\ - 7 = - 7$
also,
$\alpha \beta = \frac{c}{a} \\ ( - 8)(1) = \frac{ - 8}{1} \\ - 8 = - 8$
hence verified
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Answer 2

Hiii friend,


P(X) => X²+7X-8

=> X²+8X-X-8

=> X(X+8)-1(X+8)

=> (X+8)(X-1)

=> (X+8) = 0 OR (X-1) = 0

=> X = -8 OR X = 1


-8, 1 are the two zeros of the polynomial.

Relationship between the zeros and the coefficient.

Sum of zeros = (Alpha + Beta) = (-8+1) = 7/1 = Coefficient of X/Coefficient of X².


And,

Product of zeros = (Alpha × Beta) = -8 × 1 = -8/1 = Constant term/Coefficient of X².



HOPE IT WILL HELP YOU....... :-)