Question

plz answer it for me
those who will ans. all the question for me i will give him all the scores that is on my point board​

Answer 1

Answer:

Step-by-step explanation:

1

Number of zeroes of a polynomial refers to the number of points its

graph intersects the x axis

Here it intersects the x axis at 3 points

Yherefore the number of zeroes of the graph are 3.

2

given

$p(x)=ax^2-3(a-1)x-1\\\\given\\\\1\;is\;a\;zero\;of\;p(x)\\\\=>\;p(1)=0\\\\a-3(a-1)-1=0\\\\-2a+2=0\\\\a=1$

3

given

$\alpha+\beta =6\\\\\alpha \beta =6\\\\WKT\;polynomial\;with\;zeroes\;\alpha\;and\;\beta\;is\\\\x^2-(\alpha+ \beta)+\alpha \beta  =0\\\\x^2-6x+6=0\\\\is\;the\;required\;polynomial.$

4

given

$\alpha+\beta =-\frac{9}{2}\\\\\alpha \beta =-\frac{3}{2}\\\\WKT\;polynomial\;with\;zeroes\;\alpha\;and\;\beta\;is\\\\x^2-(\alpha+ \beta)+\alpha \beta  =0\\\\2x^2+9x-3=0\\\\is\;the\;required\;polynomial.$

5

given

$\sqrt{3}x^2-8x+4\sqrt{3}=0\\\\\sqrt{3}x^2-6x-2x+4\sqrt{3}=0\\\\\sqrt{3}x(x-2\sqrt{3})-2(x-2\sqrt{3})=0\\\\(\sqrt{3}x-2)(x-2\sqrt{3})=0\\\\x=\frac{2}{\sqrt{3}}\\\\x=2\sqrt{3}$

6

given

$x^3-4x^2-3x+12\\\\=x^2(x-4)-3(x-4)\\\\=(x^2-3)(x-4)\\\\=(x-4)(x^2-\sqrt{3}x+\sqrt{3}x-3)\\\\=(x-4)(x(x-\sqrt{3})+\sqrt{3}(x-\sqrt{3}))\\=(x-4)(x-\sqrt{3})(x+\sqrt{3})\\\\\\Therefore\;the\;third\;zero\;is\;4$

7

given

$2x^4-9x^3+5x^2+3x-1\\\\=2x^4-8x^3+2x^2-x^3+4x^2-x-x^2+4x-1\\\\=2x^2(x^2-4x+1)-x(x^2-4x+1)-1(x^2-4x+1)\\\\=(x^2-4x+1)(2x^2-x-1)\\\\=(x^2-(2+\sqrt{3})x-(2-\sqrt{3})x+(2+\sqrt{3})(2-\sqrt{3}))(2x^2-2x+x-1)\\\\=(x(x-(2+\sqrt{3})-(2-\sqrt{3})(x-(2+\sqrt{3}))(2x(x-1)+1(x-1))\\\\=(x-(2+\sqrt{3}))(x-(2-\sqrt{3}))(2x+1)(x-1)\\\\Therefore\;the \;roots\;are\\\\2+\sqrt{3}\\\\2-\sqrt{3}\\\\1\\\\-\frac{1}{2}$

8

given

$3x^4+6x^3-2x^2-10x-5\\\\3(x^4+2x^3-\frac{2}{3}x^2-\frac{10}{3}x-\frac{5}{3})\\\\=3(x^4-\frac{5}{3}x^2+2x^3-\frac{10}{3}x+x^2-\frac{5}{3})\\\\=3(x^2(x^2-\frac{5}{3})+2x(x^2-\frac{5}{3})+(x^2-\frac{5}{3}))\\\\=3(x^2-\frac{5}{3})(x^2+2x+1)\\\\=3(x-\sqrt{\frac{5}{3}})(x+\sqrt{\frac{5}{3}})(x+1)^2\\\\Therefore \;the\;other\;two\;zeroes\;are\;1\;and\;1$

9

given

$WKT\\\\Dividend=quotient\times divisor+Remainder\\\\g(x)(x-2)-2x+4=x^3-3x^2+x+2\\\\g(x)(x-2)=x^3-3x^2+3x-2\\\\g(x)(x-2)=x^3-x^2+x-2x^2+2x-2\\\\g(x)(x-2)= x(x^2-x+1)-2(x^2-x+1)\\\\g(x)(x-2)=(x-2)(x^2-x+1)\\\\g(x)=x^2-x+1$

10 and 11

given

in the attachment

HOPE IT HELPS