ddioddefn5cw

2023-02-22

${P}_{1}(x)=3{x}^{2}+10x+8\phantom{\rule{0ex}{0ex}}{P}_{2}(x)={x}^{3}+{x}^{2}+2x+t$

are two polynomials.

When one of the factors of ${P}_{1}(x)$ divides ${P}_{2}(x)$, 2 is the remainder obtained.

That factor is also a factor of the polynomial ${P}_{3}(x)=2(x+2)$

Find the value of ‘t’.

are two polynomials.

When one of the factors of ${P}_{1}(x)$ divides ${P}_{2}(x)$, 2 is the remainder obtained.

That factor is also a factor of the polynomial ${P}_{3}(x)=2(x+2)$

Find the value of ‘t’.

Haleigh Russo

Beginner2023-02-23Added 6 answers

Let the factor be (x-a).

Using factor theorem, ${P}_{3}(a)=0$

${P}_{3}(x)=2(x+2)\phantom{\rule{0ex}{0ex}}{P}_{3}(a)=2(a+2)=0\phantom{\rule{0ex}{0ex}}\Rightarrow a=-2$

So, the factor is $(x+2)$. Verifying that this is also a factor of ${P}_{1}(x)$ using factor theorem,

${P}_{1}(-2)=3\times (-2{)}^{2}+10\times (-2)+8\phantom{\rule{0ex}{0ex}}{P}_{1}(-2)=3\times 4-20+8=0$

By remainder theorem, when ${P}_{2}(x)$ is divided by (x+2), the remainder is ${P}_{2}(-2)$

${P}_{2}(-2)=(-2{)}^{3}+(-2{)}^{2}+2\times (-2)+t\phantom{\rule{0ex}{0ex}}{P}_{2}(-2)=-8+4-4+t=t-8$

Given remainder is 2. So,

${P}_{2}(-2)=2\phantom{\rule{0ex}{0ex}}\Rightarrow t-8=2\phantom{\rule{0ex}{0ex}}\Rightarrow t=10$

Using factor theorem, ${P}_{3}(a)=0$

${P}_{3}(x)=2(x+2)\phantom{\rule{0ex}{0ex}}{P}_{3}(a)=2(a+2)=0\phantom{\rule{0ex}{0ex}}\Rightarrow a=-2$

So, the factor is $(x+2)$. Verifying that this is also a factor of ${P}_{1}(x)$ using factor theorem,

${P}_{1}(-2)=3\times (-2{)}^{2}+10\times (-2)+8\phantom{\rule{0ex}{0ex}}{P}_{1}(-2)=3\times 4-20+8=0$

By remainder theorem, when ${P}_{2}(x)$ is divided by (x+2), the remainder is ${P}_{2}(-2)$

${P}_{2}(-2)=(-2{)}^{3}+(-2{)}^{2}+2\times (-2)+t\phantom{\rule{0ex}{0ex}}{P}_{2}(-2)=-8+4-4+t=t-8$

Given remainder is 2. So,

${P}_{2}(-2)=2\phantom{\rule{0ex}{0ex}}\Rightarrow t-8=2\phantom{\rule{0ex}{0ex}}\Rightarrow t=10$