If α, β are the zeros of the quadratic polynomial ax^2 + bx + c, a 0, then prove that a(x-α)(x-β) is equal to ax^2 – bx + c
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α,β are the zeros of ax²+bx+c=0
then, α+β=-(b/a) and α×β=c/a
then, a(x-α)(x-β)
=a(x²-αx-βx+αβ)
=a{x²-x(α+β)+αβ}
=a{x²-x(-b/a)+c/a}
=a(x²+bx/a+c/a)
=a(ax²+bx+c)/a
=ax²+bx+c (proved)
then, α+β=-(b/a) and α×β=c/a
then, a(x-α)(x-β)
=a(x²-αx-βx+αβ)
=a{x²-x(α+β)+αβ}
=a{x²-x(-b/a)+c/a}
=a(x²+bx/a+c/a)
=a(ax²+bx+c)/a
=ax²+bx+c (proved)
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