5 Marks Questions

Question 15 Marks

Find the remainder obtained on dividing $p(x) = x^3 + 1 by x + 1.$

Answer

By long division,we have,

Therefore, remainder is $0.$
Here $p(x) = x^3 + 1$, and the root of $x + 1 = 0$ is $x = –1.$ We have,
$p(–1) = (–1)^3 + 1$
$= –1 + 1 = 0,$
which is equal to the remainder obtained by actual division.

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Question 25 Marks

Divide the polynomial $3x^4 – 4x^3 – 3x –1$ by $x – 1$

Answer

By long division, we have:

Since the remainder is $– 5.$
Now, the zero of $x – 1$ is $1.$
Therefore, putting $x = 1$ in $p(x)$, we have,
$p(1) = 3(1)^4 – 4(1)^3 – 3(1) – 1 = 3 – 4 – 3 – 1$
$= – 5$, which is the remainder

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Question 35 Marks

Divide $p(x) by g(x)$, where $p(x) = x + 3x^2 – 1$ and $g(x) = 1 + x$

Answer

We carry out the process of division by means of the following steps:
Step $1$: We write the dividend $x +3 x ^2-1$ and the divisor $1+ x$ in the standard form, i.e., after arranging the terms in the descending order of their degrees. Therefore, the dividend is $3 x^2+x-1$ and divisor is $x+1$.
Step $2 $: We divide the first term of the dividend by the first term of the divisor, i.e., we divide $3 x^2$ by $x$, and get $3 x$.
This gives us the first term of the quotient. $\frac{3 x^2}{x}=3 x=$ first term of quotient
Step 3 : We multiply the divisor by the first term of the quotient, and subtract this product from the dividend, i.e., we multiply $x+1$ by $3 x$ and subtract the product $3 x^2+3 x$ from the dividend $3 x^2+x-1$. This gives us the remainder as $2 x-1$

Step 4 : We treat the remainder $-2 x-1$ as the new dividend. The divisor remains the same. We repeat Step $2$ to get the next term of the quotient, i.e., we divide the first term $-2 x$ of the (new) dividend by the first term $x$ of the divisor and obtain $-2$ . Therefore,$ -2$ is the second term in the quotient. $\frac{-2 x}{x}=-2=$ second term of quotient $\mid$ New Quotient $=3 x -2$
Step 5 : We multiply the divisor by the second term of the quotient and subtract the product from the dividend. That is, we multiply $x+1$ by $-2$ and subtract the product $-2 x-2$ from the dividend $-2 x-1$. This gives us $1$ as the remainder.

This process continues till the remainder is $0$ or the degree of the new dividend is less than the degree of the divisor. At this stage, this new dividend becomes the remainder and the sum of the quotients gives us the whole quotient. Step $6$ : Since the degree of the new dividend $1$ is less than degree of divisor, therefore the quotient is full $3 x-2$ and the remainder is $1$ . Let us look at what we have done in the process above as a whole:

Notice that $3x^2 + x – 1 = (x + 1) (3x – 2) + 1$

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Maths 5 Marks Questions

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