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Lesson Posted on 20/08/2023 Learn Polynomials +1

Nikhil Tamoli

I am currently pursuing B.Tech in Civil Engineering. I am a final year student and I have a keen interest...

A polynomial is a mathematical expression consisting of variables (also known as indeterminates) and coefficients, combined using the operations of addition, subtraction, and multiplication. The variables in a polynomial are raised to non-negative integer powers. In other words, a polynomial is a sum... read more

A polynomial is a mathematical expression consisting of variables (also known as indeterminates) and coefficients, combined using the operations of addition, subtraction, and multiplication. The variables in a polynomial are raised to non-negative integer powers. In other words, a polynomial is a sum of terms, where each term consists of a coefficient multiplied by a variable raised to a non-negative integer exponent.

Polynomials can have various degrees. Here are some examples of polynomials of different degrees:

1. Linear Polynomial:  The degree is 1.
2. Quadratic Polynomial:P(x)=ax2+bx+c, where a, b, and c are constants. The degree is 2.
3. Cubic Polynomial: P(x)=ax3+bx2+cx+d, where a, b, c, and d are constants. The degree is 3.
4. Quartic Polynomial: P(x)=ax4+bx3+cx2+dx+e, where a, b, c, d, and e are constants. The degree is 4.

Polynomials are used extensively in mathematics and various scientific and engineering fields for modeling and solving equations, approximating functions, and more.

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Lesson Posted on 20/04/2022 Learn Polynomials

Vikas Mehra

I am an experienced ,qualified teacher with over 4 years of experience in teaching mathematics and economic....

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Lesson Posted on 19/04/2022 Learn Polynomials

Vikas Mehra

I am an experienced ,qualified teacher with over 4 years of experience in teaching mathematics and economic....

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Answered on 16 Apr Learn Polynomials

If 11 is a zero of the polynomial p(x)=ax2−3(a−1)x−1 read more

If 11 is a zero of the polynomial p(x)=ax2−3(a−1)x−1$Substituting&space;$$x&space;=&space;1$$&space;into&space;the&space;polynomial&space;equation:&space;$&space;p(1)&space;=&space;a(1)^2&space;-&space;3(a&space;-&space;1)(1)&space;-&space;1&space;=&space;0&space;$&space;$&space;a&space;-&space;3(a&space;-&space;1)&space;-&space;1&space;=&space;0&space;$&space;Now,&space;let's&space;solve&space;for&space;$$a$$:&space;$&space;a&space;-&space;3a&space;+&space;3&space;-&space;1&space;=&space;0&space;$&space;$&space;-2a&space;+&space;2&space;=&space;0&space;$&space;$&space;-2a&space;=&space;-2&space;$&space;$&space;a&space;=&space;\frac{-2}{-2}&space;$&space;$&space;a&space;=&space;1&space;$&space;So,&space;the&space;value&space;of&space;$$a$$&space;for&space;which&space;$$1$$&space;is&space;a&space;zero&space;of&space;the&space;polynomial&space;$$p(x)&space;=&space;ax^2&space;-&space;3(a&space;-&space;1)x&space;-&space;1$$&space;is&space;$$a&space;=&space;1$$.$

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Answered on 16 Apr Learn Polynomials

To find aa if (x+a) is a factor of read more

To find aa if (x+a) is a factor of$$$2x^2&space;+&space;2ax&space;+&space;5x&space;+&space;10$$,&space;we'll&space;perform&space;polynomial&space;division.&space;Given&space;that&space;$$(x&space;+&space;a)$$&space;is&space;a&space;factor,&space;we&space;have:&space;$2x^2&space;+&space;2ax&space;+&space;5x&space;+&space;10&space;=&space;(x&space;+&space;a)(2x&space;+&space;b)$&space;Expanding&space;$$(x&space;+&space;a)(2x&space;+&space;b)$$,&space;we&space;get:&space;$2x^2&space;+&space;(2a&space;+&space;b)x&space;+&space;ab$&space;Comparing&space;coefficients&space;of&space;corresponding&space;terms&space;in&space;$$2x^2&space;+&space;2ax&space;+&space;5x&space;+&space;10$$&space;and&space;$$2x^2&space;+&space;(2a&space;+&space;b)x&space;+&space;ab$$,&space;we&space;have:&space;For&space;$$x$$&space;terms:&space;$2ax&space;+&space;5x&space;=&space;(2a&space;+&space;b)x$&space;So,&space;$$2a&space;+&space;b&space;=&space;5$$&space;For&space;constant&space;terms:&space;$10&space;=&space;ab$&space;Given&space;$$2a&space;+&space;b&space;=&space;5$$,&space;we&space;can&space;express&space;$$b$$&space;in&space;terms&space;of&space;$$a$$:&space;$b&space;=&space;5&space;-&space;2a$&space;Substituting&space;this&space;expression&space;for&space;$$b$$&space;into&space;$$10&space;=&space;ab$$,&space;we&space;get:&space;$10&space;=&space;a(5&space;-&space;2a)$&space;Expanding&space;and&space;rearranging,&space;we&space;have:&space;$10&space;=&space;5a&space;-&space;2a^2$&space;$2a^2&space;-&space;5a&space;+&space;10&space;=&space;0$&space;This&space;is&space;a&space;quadratic&space;equation.&space;We&space;can&space;solve&space;it&space;using&space;the&space;quadratic&space;formula:$

$$a&space;=&space;\frac{-b&space;\pm&space;\sqrt{b^2&space;-&space;4ac}}{2a}$&space;For&space;$$2a^2&space;-&space;5a&space;+&space;10&space;=&space;0$$,&space;we&space;have&space;$$a&space;=&space;2$$,&space;$$b&space;=&space;-5$$,&space;and&space;$$c&space;=&space;10$$.&space;Substituting&space;these&space;values&space;into&space;the&space;quadratic&space;formula:&space;$a&space;=&space;\frac{-(-5)&space;\pm&space;\sqrt{(-5)^2&space;-&space;4&space;\cdot&space;2&space;\cdot&space;10}}{2&space;\cdot&space;2}$&space;$a&space;=&space;\frac{5&space;\pm&space;\sqrt{25&space;-&space;80}}{4}$&space;$a&space;=&space;\frac{5&space;\pm&space;\sqrt{-55}}{4}$&space;Since&space;$$\sqrt{-55}$$&space;is&space;imaginary,&space;the&space;solutions&space;for&space;$$a$$&space;are&space;complex.&space;Thus,&space;there's&space;no&space;real&space;value&space;of&space;$$a$$&space;for&space;which&space;$$(x&space;+&space;a)$$&space;is&space;a&space;factor&space;of&space;$$2x^2&space;+&space;2ax&space;+&space;5x&space;+&space;10$$.$

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Answered on 16 Apr Learn Polynomials

To find the zeroes of the polynomial read more

To find the zeroes of the polynomial$T&space;$$x^2&space;+&space;2x&space;+&space;1$$,&space;we&space;need&space;to&space;solve&space;for&space;$$x$$&space;when&space;the&space;polynomial&space;equals&space;$$0$$.&space;The&space;polynomial&space;$$x^2&space;+&space;2x&space;+&space;1$$&space;can&space;be&space;factored&space;as&space;$$(x&space;+&space;1)(x&space;+&space;1)$$&space;or&space;$$(x&space;+&space;1)^2$$.&space;Setting&space;the&space;polynomial&space;equal&space;to&space;$$0$$,&space;we&space;have:&space;$(x&space;+&space;1)^2&space;=&space;0$&space;Since&space;$$(x&space;+&space;1)^2$$&space;equals&space;$$0$$,&space;$$x&space;+&space;1$$&space;must&space;equal&space;$$0$$.&space;Therefore,&space;the&space;only&space;zero&space;of&space;the&space;polynomial&space;is&space;$$x&space;=&space;-1$$.&space;So,&space;the&space;zeroes&space;of&space;the&space;polynomial&space;$$x^2&space;+&space;2x&space;+&space;1$$&space;is&space;$$x&space;=&space;-1$$.$

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Answered on 16 Apr Learn Polynomials

To find the zeroes of the polynomial read more

To find the zeroes of the polynomial $T&space;$$x^2&space;-&space;x&space;-&space;6$$,&space;we&space;need&space;to&space;solve&space;for&space;$$x$$&space;when&space;the&space;polynomial&space;equals&space;$$0$$.&space;We&space;can&space;factor&space;the&space;polynomial&space;as&space;follows:&space;$x^2&space;-&space;x&space;-&space;6&space;=&space;(x&space;-&space;3)(x&space;+&space;2)$&space;Setting&space;the&space;polynomial&space;equal&space;to&space;$$0$$,&space;we&space;have:&space;$(x&space;-&space;3)(x&space;+&space;2)&space;=&space;0$&space;Now,&space;we&space;solve&space;for&space;$$x$$&space;by&space;setting&space;each&space;factor&space;equal&space;to&space;$$0$$:&space;1.&space;$$x&space;-&space;3&space;=&space;0$$&space;$x&space;=&space;3$&space;2.&space;$$x&space;+&space;2&space;=&space;0$$&space;$x&space;=&space;-2$&space;So,&space;the&space;zeroes&space;of&space;the&space;polynomial&space;$$x^2&space;-&space;x&space;-&space;6$$&space;are&space;$$x&space;=&space;3$$&space;and&space;$$x&space;=&space;-2$$.$

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Answered on 16 Apr Learn Polynomials

Let's denote the zeroes of the quadratic polynomial as$$$x_1$$&space;and&space;$$x_2$$.&space;According&space;to&space;the&space;given&space;conditions:&space;1.&space;The&space;sum&space;of&space;the&space;zeroes&space;$$x_1&space;+&space;x_2&space;=&space;3$$.&space;2.&space;The&space;product&space;of&space;the&space;zeroes&space;$$x_1&space;\cdot&space;x_2&space;=&space;-2$$.&space;We&space;know&space;that&space;for&space;a&space;quadratic&space;polynomial&space;$$ax^2&space;+&space;bx&space;+&space;c$$,&space;the&space;sum&space;of&space;the&space;zeroes&space;is&space;given&space;by&space;$$-\frac{b}{a}$$&space;and&space;the&space;product&space;of&space;the&space;zeroes&space;is&space;given&space;by&space;$$\frac{c}{a}$$.&space;Given&space;that&space;$$x_1&space;+&space;x_2&space;=&space;3$$&space;and&space;$$x_1&space;\cdot&space;x_2&space;=&space;-2$$,&space;we&space;can&space;set&space;up&space;the&space;following&space;system&space;of&space;equations:$

$1.&space;$$x_1&space;+&space;x_2&space;=&space;3$$&space;2.&space;$$x_1&space;\cdot&space;x_2&space;=&space;-2$$&space;We&space;can&space;rewrite&space;the&space;first&space;equation&space;as&space;$$x_1&space;+&space;x_2&space;-&space;3&space;=&space;0$$&space;and&space;the&space;second&space;equation&space;as&space;$$x_1&space;\cdot&space;x_2&space;+&space;2&space;=&space;0$$.&space;Since&space;the&space;sum&space;and&space;product&space;of&space;the&space;zeroes&space;of&space;the&space;quadratic&space;polynomial&space;are&space;$$3$$&space;and&space;$$-2$$&space;respectively,&space;we&space;can&space;express&space;the&space;quadratic&space;polynomial&space;as:&space;$p(x)&space;=&space;(x&space;-&space;x_1)(x&space;-&space;x_2)$&space;Substituting&space;the&space;values&space;of&space;$$x_1$$&space;and&space;$$x_2$$&space;from&space;the&space;given&space;conditions&space;into&space;the&space;expression&space;for&space;the&space;quadratic&space;polynomial:&space;$p(x)&space;=&space;(x&space;-&space;3)(x&space;+&space;2)$&space;Expanding&space;this&space;expression,&space;we&space;get:&space;$p(x)&space;=&space;x^2&space;-&space;x&space;-&space;6$&space;So,&space;the&space;quadratic&space;polynomial&space;satisfying&space;the&space;given&space;conditions&space;is&space;$$x^2&space;-&space;x&space;-&space;6$$.$

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Answered on 16 Apr Learn Polynomials

We are given that : : read more

We are given that : :$$$&space;\alpha&space;$$&space;and&space;$$&space;eta&space;$$&space;are&space;the&space;zeroes&space;of&space;the&space;quadratic&space;polynomial&space;$$&space;x^2&space;-&space;6x&space;+&space;a&space;$$.&space;By&space;Vieta's&space;formulas,&space;for&space;a&space;quadratic&space;equation&space;$$&space;ax^2&space;+&space;bx&space;+&space;c&space;=&space;0&space;$$,&space;the&space;sum&space;of&space;the&space;roots&space;($$&space;\alpha&space;+&space;eta&space;$$)&space;is&space;equal&space;to&space;$$&space;-\frac{b}{a}&space;$$&space;and&space;the&space;product&space;of&space;the&space;roots&space;($$&space;\alpha&space;\times&space;eta&space;$$)&space;is&space;equal&space;to&space;$$&space;\frac{c}{a}&space;$$.&space;So,&space;we&space;have:&space;1.&space;$$&space;\alpha&space;+&space;eta&space;=&space;6&space;$$&space;2.&space;$$&space;\alpha&space;\times&space;eta&space;=&space;a&space;$$&space;Given&space;that&space;$$&space;3\alpha&space;+&space;2eta&space;=&space;20&space;$$,&space;we&space;can&space;express&space;this&space;equation&space;in&space;terms&space;of&space;$$&space;\alpha&space;$$&space;and&space;$$&space;eta&space;$$:&space;$&space;3\alpha&space;+&space;2eta&space;=&space;20&space;$&space;Now,&space;let's&space;use&space;the&space;values&space;of&space;$$&space;\alpha&space;$$&space;and&space;$$&space;eta&space;$$&space;from&space;the&space;given&space;conditions&space;to&space;find&space;$$&space;a&space;$$:&space;$&space;3\alpha&space;+&space;2eta&space;=&space;3(\alpha)&space;+&space;2(eta)&space;=&space;20&space;$&space;Substituting&space;$$&space;\alpha&space;+&space;eta&space;=&space;6&space;$$&space;into&space;the&space;equation:&space;$&space;3(6)&space;=&space;20&space;$&space;$&space;18&space;=&space;20&space;$$

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Answered on 16 Apr Learn Polynomials

To find the quadratic polynomial whose zeroes are $$$1$$&space;and&space;$$-3$$,&space;we&space;can&space;use&space;the&space;fact&space;that&space;if&space;$$&space;\alpha&space;$$&space;and&space;$$&space;eta&space;$$&space;are&space;the&space;zeroes&space;of&space;a&space;quadratic&space;polynomial&space;$$&space;ax^2&space;+&space;bx&space;+&space;c&space;$$,&space;then&space;the&space;polynomial&space;can&space;be&space;expressed&space;as&space;$$&space;a(x&space;-&space;\alpha)(x&space;-&space;eta)&space;$$.&space;Given&space;that&space;the&space;zeroes&space;are&space;$$1$$&space;and&space;$$-3$$,&space;we&space;can&space;write&space;the&space;quadratic&space;polynomial&space;as:&space;$&space;p(x)&space;=&space;a(x&space;-&space;1)(x&space;-&space;(-3))&space;$&space;$&space;p(x)&space;=&space;a(x&space;-&space;1)(x&space;+&space;3)&space;$&space;Expanding&space;this&space;expression,&space;we&space;get:&space;$&space;p(x)&space;=&space;a(x^2&space;+&space;3x&space;-&space;x&space;-&space;3)&space;$&space;$&space;p(x)&space;=&space;a(x^2&space;+&space;2x&space;-&space;3)&space;$&space;Now,&space;to&space;verify&space;the&space;relation&space;between&space;the&space;coefficients&space;and&space;the&space;zeroes&space;of&space;the&space;polynomial,&space;let's&space;find&space;the&space;values&space;of&space;$$a$$&space;using&space;one&space;of&space;the&space;coefficients.$

$Given&space;that&space;$$1$$&space;and&space;$$-3$$&space;are&space;the&space;zeroes&space;of&space;the&space;polynomial,&space;if&space;we&space;substitute&space;$$x&space;=&space;1$$&space;into&space;$$p(x)$$,&space;it&space;should&space;equal&space;$$0$$,&space;and&space;similarly,&space;if&space;we&space;substitute&space;$$x&space;=&space;-3$$,&space;$$p(x)$$&space;should&space;also&space;equal&space;$$0$$.&space;Let's&space;substitute&space;$$x&space;=&space;1$$&space;into&space;$$p(x)$$:&space;$&space;p(1)&space;=&space;a(1^2&space;+&space;2(1)&space;-&space;3)&space;$&space;$&space;p(1)&space;=&space;a(1&space;+&space;2&space;-&space;3)&space;$&space;$&space;p(1)&space;=&space;a(0)&space;$&space;$&space;p(1)&space;=&space;0&space;$&space;Now,&space;let's&space;substitute&space;$$x&space;=&space;-3$$&space;into&space;$$p(x)$$:&space;$&space;p(-3)&space;=&space;a((-3)^2&space;+&space;2(-3)&space;-&space;3)&space;$&space;$&space;p(-3)&space;=&space;a(9&space;-&space;6&space;-&space;3)&space;$&space;$&space;p(-3)&space;=&space;a(0)&space;$&space;$&space;p(-3)&space;=&space;0&space;$&space;Since&space;$$p(1)&space;=&space;0$$&space;and&space;$$p(-3)&space;=&space;0$$,&space;we&space;have&space;verified&space;that&space;$$1$$&space;and&space;$$-3$$&space;are&space;indeed&space;the&space;zeroes&space;of&space;the&space;polynomial&space;$$p(x)$$.&space;Thus,&space;the&space;quadratic&space;polynomial&space;whose&space;zeroes&space;are&space;$$1$$&space;and&space;$$-3$$&space;is&space;$$p(x)&space;=&space;a(x^2&space;+&space;2x&space;-&space;3)$$.$

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