Polynomials (1 Viewer)

cos16mh · Jul 12, 2018

Help please

Polynomial P(x) = ax^3 - bx^2 +cx -8 has zeros at 2 and -1 and P(3) =28 Find the values of a,b and c

fan96 · Jul 13, 2018

By the Factor Theorem, $P(2) = 0$ and $P(-1) = 0$ .

$\begin{aligned} P(3) = 28 &\implies a(\phantom{-}3)^3-b(\phantom{-}3)^2+c(\phantom{-}3)-8 = 28 \\ P(2) = \phantom{-}0 &\implies a(\phantom{-}2)^3-b(\phantom{-}2)^2+c(\phantom{-}2)-8= 0 \\ P(-1) = \phantom{0}0 &\implies a(-1)^3-b(-1)^2+c(-1)-8 = 0\end{aligned}$ .

Solving these equations will give you the values of $a,\,b,\,c$ .

Drongoski · Jul 13, 2018

But a 2U student may not know how to solve a system of 3 linear equations in 3 unknowns.

HoldingOn · Jul 13, 2018

Polynomials is an Extension 1 topic anyway.

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Polynomials (1 Viewer)

cos16mh

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fan96

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Drongoski

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HoldingOn

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