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2023-02-22

${P}_{1}\left(x\right)=3{x}^{2}+10x+8\phantom{\rule{0ex}{0ex}}{P}_{2}\left(x\right)={x}^{3}+{x}^{2}+2x+t$
are two polynomials.
When one of the factors of ${P}_{1}\left(x\right)$ divides ${P}_{2}\left(x\right)$, 2 is the remainder obtained.
That factor is also a factor of the polynomial ${P}_{3}\left(x\right)=2\left(x+2\right)$
Find the value of ‘t’.

Haleigh Russo

Let the factor be (x-a).
Using factor theorem, ${P}_{3}\left(a\right)=0$
${P}_{3}\left(x\right)=2\left(x+2\right)\phantom{\rule{0ex}{0ex}}{P}_{3}\left(a\right)=2\left(a+2\right)=0\phantom{\rule{0ex}{0ex}}⇒a=-2$
So, the factor is $\left(x+2\right)$. Verifying that this is also a factor of ${P}_{1}\left(x\right)$ using factor theorem,
${P}_{1}\left(-2\right)=3×\left(-2{\right)}^{2}+10×\left(-2\right)+8\phantom{\rule{0ex}{0ex}}{P}_{1}\left(-2\right)=3×4-20+8=0$
By remainder theorem, when ${P}_{2}\left(x\right)$ is divided by (x+2), the remainder is ${P}_{2}\left(-2\right)$
${P}_{2}\left(-2\right)=\left(-2{\right)}^{3}+\left(-2{\right)}^{2}+2×\left(-2\right)+t\phantom{\rule{0ex}{0ex}}{P}_{2}\left(-2\right)=-8+4-4+t=t-8$
Given remainder is 2. So,
${P}_{2}\left(-2\right)=2\phantom{\rule{0ex}{0ex}}⇒t-8=2\phantom{\rule{0ex}{0ex}}⇒t=10$

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