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Multiplying Polynomials|Dividing Polynomials





Multiplying polynomials


Multiplication of polynomial is simple task and we can follow below steps to multiply polynomials

  1. Arrange both the polynomial is same order of exponent . It would be good to have terms arrange from highest exponent to lowest exponent
  2. We have to distribute each term of the first polynomial to every term of the second polynomial.
    a. when you multiply two terms together you must multiply the coefficient (numbers) and add the exponents
    b. Also as we already know ++ equals =, +- or -+ equals - and -- equals +
  3. group like terms
multiplying polynomials

So multiplication may involve
  • Multiplication of monomial to monomial
  • Multiplication of monomial to binomial, trinomial or more terms polynomials
  • Multiplication of binomial, trinomial or more terms polynomials to monomial
  • Multiplication of binomial to binomial, trinomial or more terms polynomials
  • Multiplication of trinomial to trinomial or more terms polynomials
Multiple the Monomials
  1. $x^2 \times (2x^{22}) \times (4x^26)$
  2. $2 (\frac {-10xy^3}{3}) \times (\frac {6x^3y}{5})$
  3. $( x) \times (x^2) \times (x^3) \times (x^8)$
Answer:
We will use the below property extensively in above questions
$x^m \times x^n \times x^o=x^{m+n+o}$
1. As you know from above
So, we get
$x^2 \times (2x^{22}) \times (4x^{26}) =8a^{48}$
2. $2 (\frac {-10xy^3}{3}) \times (\frac {6x^3y}{5})$ $=(-8x^4y^4)$
3.$( x) \times (x^2) \times (x^3) \times (x^8)$
$=x^{14}$
Multiply the binomials.
(i) (2x + 5) and (4x - 3)
(ii) (x - 8) and (3x - 4)
(iii)(2.5x - 0.5) and (2.5x + 0.5)
Answer: 
Let  $( a+b) (c+d)$ to be done
then
$( a+b) (c+d)= a(c+d) + b( c+d)$
$=(a \times c)+(a \times d)+(b \times c)+(b \times d)$
The above expression of multiplying binomials is called FOIL method
FOIL stands FIRST ($a \times c$) OUTER ($a \times d$) INNER ($b \times c $) LAST($b \times d$)
We will use the same concept in all the question below
1. $(2x + 5)(4x - 3)$
$=2x \times 4x - 2x \times 3 +5 \times 4x- 5 \times 3$
$=8x^2 - 6x + 20x -15$
$=8x^2 + 14x -15$
2. $( y - 8)(3y - 4)$
$= y \times 3y - 4y - 8 \times 3y + 32$
$= 3y^2- 4y - 24y + 32$
$= 3y^2- 28y + 32$
3. $(2.5l - 0.5m)(2.5l + 0.5)$
Using $(a+b)(a-b) = a^2- b^2$
We get ,
$= 6.25l^2- 0.25m^2$
Multiply Binomial by Trinomial
$(3x + 2)(4x^2 - 7x + 5)$
Answer
Let  $( a+b) (c+d+e)$ to be done
then
$( a+b) (c+d+e)= a(c+d+e) + b( c+d+e)$
$=(a \times c)+(a \times d)+ (a \times e)+ (b \times c)+(b \times d) +(b \times e) $
We will use the same concept in all the question below
$=3x(4x^2 - 7x + 5) + 2(4x^2 - 7x + 5)$
$=12x^3 -21x^2 +15x +8x^2 -14x+10$
$=12x^3 -13x^2 +x+10$
 
Practice Questions
  • $(x-5)(x-9)$
  • $\left(x + 10\right) \left(x^{2} + x + 2\right)$
  • $3x^2 \times 4x^{20}$
  • $\left(x + 4\right) \left(x^{2} + 4 x + 7\right)$
  • $(x+9)(x-6)$
  • $\left(x + 2\right) \left(x + 4\right)^{2}$
  • $\left(x + 8\right) \left(x^{2} + 2 x + 9\right)$
  • $(x-2)(x-2)$

Dividing Polynomials


When a polynomial p(x) is divided by the polynomial g(x), we get quotient q(x) and remainder r(x)

$p(x)=g(x).q(x)+r(x)$
Notes
1. The degree of the reminder r(x) is always less then divisor g(x)

Now Let us see how to divide the polynomial by another non-zero polynomial

Steps to divide a polynomial by another polynomial. This is also called the long division method of polynomials

1. Arrange the term in decreasing order in both the polynomial
2. Divide the highest degree term of the dividend by the highest degree term of the divisor to obtain the first term
3. Now We multiply the divisor by the first term of the quotient, and subtract this product from the dividend
4. Similar steps are followed till we get the reminder whose degree is less than of divisor
Example
Divide p(x) by g(x), where $p(x) = x + 4x^2 -1$ and $g(x) = 1 + x$
Solution
We carry out the process of division by means of the following steps:
Step 1 We write the dividend $x + 4x^2 - 1$ and the divisor 1 + x. in the standard form, i.e., after arranging the terms in the descending order of their degrees. So, the dividend is $4x^2 + x -1$ and divisor is x + 1.
Step 2We divide the first term of the dividend by the first term of the divisor, i.e., we divide $4x^2$ by x, and get 4x. This gives us the first term of the quotient.
Step 3We multiply the divisor by the first term of the quotient, and subtract this product from the dividend, i.e., we multiply $x + 1$ by 4x and subtract the product $4x^2 + 4x$ from the dividend $4x^2 + x - 1$. This gives us the remainder as -3x - 1.
Step 4 We treat the remainder -3x - 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 -3x of the (new) dividend by the first term x of the divisor and obtain -3. Thus, -3 is the second term in the quotient.
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 - 3 and subtract the product -3x - 3 from the dividend - 3x - 1. This gives us 2 as the remainder

Now the degree of the reminder is less than degree of the divisor so process stops here

Complete division is illustrated below
dividing polynomials

Quotient is 3x -3 and Reminder is 2
Now $4x^2 + x -1= (x+1)(4x-3) +2$
Practice Questions
  • Divide $x^{3} + 12 x^{2} + 39 x + 29$ by $(x+1)$
  • Divide $x^{3} + 14 x^{2} + 43 x + 32$ by $(x+3)$
  • Divide $x^{3} + 5 x^{2} + 6 x + 2$ by $(x+1)$
  • Divide $x^{2} + 2 x - 48$ by $(x+8)$
  • Divide $x^{3} + 12 x^{2} + 47 x + 60$ by $(x+3)$

Remainder Theorem's

If p(x) is an polynomial of degree greater than or equal to 1 and p(x) is divided by the expression (x-a),then the reminder will be p(a)

Important notes
  1. for (x-a) then remainder P(a)
  2. for (x+a) => x -(-a),then remainder will be P(-a)
  3. for (ax-b) => a(x- b/a) ,the remainder will be P(b/a)
  4. for (ax+b) => a(x+b/a),the remainder will be P(-b/a)
  5. for (b-ax)=> -a(x-b/a),the remainder will be P(b/a)

    Quiz Time

    Question 1 $(x+2)(x+8)$ ?
    A. $x^2 +16$
    B. $x^2 +16+10x$
    C. $x^2 +16+2x$
    D. $x^2 +16+8x$
    Question 2 $x^2 \times x^3 \times x^{11}$ ?
    A. $x^{16}$
    B. $x^{17}$
    C. $x^{66}$
    D. $66x$
    Question 3 when $x^{2} - 5 x - 14 $ divide by $(x+2)$
    A. Remainder is (x-7)
    B. Quotient is (x+7)
    C. Remainder is 0
    D. None of these
    Question 4 $\left(x + 9\right) \left(x^{2} + 4 x + 5\right)$
    A. $x^{3} + 13 x^{2} - 41 x + 45$
    B. $x^{3} + 12 x^{2} + 41 x + 45$
    C. $x^{3} + 13 x^{2} + 41 x - 45$
    D. $x^{3} + 13 x^{2} + 41 x + 45$
    Question 5 $\left(x + 1\right) \left(x + 4\right)^{2}$
    A. $x^{3} - 9 x^{2} + 24 x + 16$
    B. $x^{3} + 9 x^{2} + 24 x - 16$
    C. $x^{3} + 9 x^{2} + 24 x + 16$
    D. none of the above
    Question 6 $x(x^5+11)$ ?
    A. $x^6+11x$
    B. $x^6+11$
    C. $x^5+11x$
    D. none of the above

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    Also Read



    Reference Books for class 9 Math

    Given below are the links of some of the reference books for class 9 Math.

    1. Mathematics for Class 9 by R D Sharma One of the best book for studying class 9 level mathematics. It has lot of problems to be solved.
    2. Secondary School Mathematics for Class 9 by R S Aggarwal This is also as good as R.D. Sharma. Either this or the book by R.D. Sharma will do. I find book R.S. Aggarwal little bit more challenging than the one by R.D. Sharma.
    3. Pearson IIT Foundation Series - Maths - Class 9 Buy this book if you want to challenge yourself further and want to prepare for JEE foundation.
    4. Pearson IIT Foundation Physics, Chemistry & Maths combo for Class 9 Only buy if you are prepared to study extra topics and want to take your studies a step further. You might need help to understand topics in these books.

    You can use above books for extra knowledge and practicing different questions.



    Class 9 Maths Class 9 Science





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