Consider the function \(P(x)=(x-1)^2(x+2)\left(x^2+3 x-4\right)\)
- By expressing \(P(x)\) as a product of its linear factors, identify its zeroes and the multiplicity of each zero. (2 marks)
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- Without using calculus, draw a sketch of \(y=P(x)\) showing any \(x\)-intercepts. (2 marks)
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a. \(P(x)=(x-1)^3(x+2)(x+4)\)
\(\text{Roots:}\)
\(x=-2 \ \ \text{(multiplicity 1)}\)
\(x=-4 \ \ \text{(multiplicity 1)}\)
\(x=1 \ \ \text{(multiplicity 3)}\)
b.
Show Worked Solution
| a. | \(P(x)\) | \(=(x-1)^2(x+2)\left(x^2+3 x-4\right)\) |
| \(=(x-1)^2(x+2)(x-1)(x+4)\) | ||
| \(=(x-1)^3(x+2)(x+4)\) |
\(\text{Roots:}\)
\(x=-2 \ \ \text{(multiplicity 1)}\)
\(x=-4 \ \ \text{(multiplicity 1)}\)
\(x=1 \ \ \text{(multiplicity 3)}\)
b. \(\text{Degree} \ P(x)=5, \ \ \text {Leading coefficient }=1\)
\(\text{As} \ \ x \rightarrow \infty, \ y \rightarrow \infty\)
\(\text{As} \ \ x \rightarrow -\infty, \ y \rightarrow -\infty\)
\(\text{At} \ \ x=0, y=-8\)
