Question

Polynomials p(x)=4x*x*x-2x*x+px+5 and q(x)=x*x*x+6x*x+p, leave the remainders a and b respectively, when divided by (x-2). Find the value of people, if a+b =0

Answer 1

Answer on Question #73398 – Math – Algebra

Question

Polynomials $p(x) = 4x^3 - 2x^2 + px + 5$ and $p(x) = x^3 + 6x^2 + p$, leave the remainders $a$ and $b$ respectively, when divided by $(x - 2)$. Find the value of $p$, if $a + b = 0$.

Solution

The Remainder Theorem

Let $P(x)$ be any polynomial of degree greater than or equal one and let $c$ be any real number. If $P(x)$ is divided by the linear polynomial $(x - c)$, then the remainder is $P(c)$.

If $p(x)$ is divided by the linear polynomial $(x - 2)$, then the remainder is $p(2)$

remainder = $p(2) = 4(2)^3 - 2(2)^2 + p(2) + 5 = 2p + 29$

We have that $2p + 29 = a$

If $q(x)$ is divided by the linear polynomial $(x - 2)$, then the remainder is $q(2)$

remainder = $q(2) = (2)^3 + 6(2)^2 + p = p + 32$

We have that $p + 32 = b$

If $a + b = 0$, then

$$2p + 29 + p + 32 = 0$$ $$3p = -61$$ $$p = -\frac{61}{3}$$

Check

$$p(x) = 4x^3 - 2x^2 - \frac{61}{3}x + 5$$ $$\frac{p(x)}{x - 2} = \frac{4x^3 - 2x^2 - \frac{61}{3}x + 5}{x - 2} = 4x^2 + 6x - \frac{25}{3} + \frac{-\frac{35}{3}}{x - 2}$$ $$q(x) = x^3 + 6x^2 - \frac{61}{3}$$ $$\frac{q(x)}{x - 2} = \frac{x^3 + 6x^2 - \frac{61}{3}}{x - 2} = x^2 + 8x + 16 + \frac{\frac{35}{3}}{x - 2}$$ $$-\frac{35}{3} + \frac{35}{3} = 0$$

Answer: $p = -\frac{61}{3}$.

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