Factorize a cubic function and identify whether the three roots are all integers

Question Sample Titled 'Factorize a cubic function and identify whether the three roots are all integers'

Let ${f{{\left({x}\right)}}}={\left({x}-{3}\right)}^{{2}}{\left({x}+{h}\right)}+{k}$ , where ${h}$ and ${k}$ are constants. When ${f{{\left({x}\right)}}}$ is divided by ${x}-{3}$ , the remainder is ${4}$. It is given that ${f{{\left({x}\right)}}}$ is divisible by ${x}-{4}$.

 (a) Find ${h}$ and ${k}$ . (3 marks) (b) Someone claims that all the roots of the equation ${f{{\left({x}\right)}}}$ $={0}$ are integers. Do you agree? Explain your answer. (3 marks)

 (a) By remainder theorem, ${f{{\left({3}\right)}}}$ $={4}$ ${\left({3}-{3}\right)}^{{2}}{\left({x}+{h}\right)}+{k}$ $={4}$ ${k}$ $={4}$ 1A  By remainder theorem, ${f{{\left({4}\right)}}}$ $={0}$ 1M ${\left({4}-{3}\right)}^{{2}}{\left({4}+{h}\right)}+{4}$ $={0}$ ${h}$ $=-{8}$ 1A  (b) ${f{{\left({x}\right)}}}={\left({x}-{3}\right)}^{{2}}{\left({x}-{8}\right)}+{4}$ by part(a) ${f{{\left({x}\right)}}}$ $={0}$ ${\left({x}-{3}\right)}^{{2}}{\left({x}-{8}\right)}+{4}$ $={0}$ ${\left({x}^{{2}}-{6}{x}+{9}\right)}{\left({x}-{8}\right)}+{4}$ $={0}$ ${\left({x}^{{3}}-{6}{x}^{{2}}+{9}{x}\right)}+{\left(-{8}{x}^{{2}}+{48}{x}-{72}\right)}+{4}$ $={0}$ ${x}^{{3}}-{14}{x}^{{2}}+{57}{x}-{68}$ $={0}$ 1A By long division, ${\left({x}-{4}\right)}{\left({x}^{{2}}-{10}{x}+{17}\right)}$ $={0}$ 1M for ${\left({x}-{4}\right)}{\left({a}{x}^{{2}}+{b}{x}+{c}\right)}$, steps need not to be shown ${\left({x}-{4}\right)}{\left[{x}+{\left({5}-{2}\sqrt{{{2}}}\right)}\right]}{\left[{x}-{\left({5}-{2}\sqrt{{{2}}}\right)}\right]}$ $={0}$ ∴  ${x}={4}$ or ${x}={5}\pm{2}\sqrt{{{2}}}$ Note that both ${5}-{2}\sqrt{{{2}}}$ and ${5}+{2}\sqrt{{{2}}}$ are not integers. Thus, the claim is diagreed. 1A f.t.

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