Question

What is the remainder, if the polynomial
${x}^{3} + 4{x}^{2} + 4x - 3$
is divided by
$x - 1$
​

Answer 1

Let p(x) = ax3 + 4x2 + 3x – 4 and q(x) = x3 – 4x + a be the given  polynomials.

When p(x) and q(x) are divided by (x – 3) the remainder are p(3)  and q(3)  respectively.

p(3) = q(3)  given

a(3)3 + 4(3)2 + 3 × 3 – 4 = 33 – 4 × 3 + a

⇒  27a + 36 + 9 – 4 = 27 – 12 + a

⇒ 26a = 15 – 41

⇒ 26a = - 26

∴    a = - 26/26 = - 1

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Answer 2

$\underline{ \sf\large{Solution- }}$

Solving by Remainder Theorem :

Here,

$\tt \implies \: p(x) = {x}^{3} + 4{x}^{2} +4x - 3$

• Zero of x-1 is 1

So,

$\tt \implies \: {(1)}^{3} + 4{(1)}^{2} + 4(1) - 3$

$\tt \implies \: 1 + 4 + 1$

$\tt \implies \: 6$

Therefore,

• We got the remainder as 6.

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⟹ Remainder Theorem : Let p(x) be any polynomial of degree greater than or equal to one and let a be any real number. If p(x) is divided by the linear equation x-a, then the remainder is p(a).