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Factoring with Algebra Tiles


Represent the quadratic polynomial 2x2 + x – 6 using algebra tiles and determine the equivalent factored form.
The number of zero pairs needed to model this polynomial is .
The equivalent factored form is .

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HomertheGenius
We have to determine the equivalent factored form for the quadratic polynomial:
2 x² + x - 6
2 x² + x - 6 = 2 x² + 4 x - 3 x - 6 =
= 2 x ( x + 2 ) - 3 ( x + 2 ) =
= ( x + 2 ) ( 2 x - 3 ).
Zero pairs:
x + 2 = 0
x = - 2;
2 x - 3 = 0
2 x =  3
x =  3 /2
Answer:
The number of zero pairs needed to model this polynomial is : two ( x = - 2 and x = 3/2 ).
The equivalent factored form is: ( x + 2 ) ( 2 x - 3 ).
rahulsharma789888

2 zero pairs needed to model this polynomial .

The equivalent factored form is

[tex]\rm\bold{(x-3/2)(x+2) =0}[/tex]

Given expression

[tex]\rm y = 2x^2+x -6.........(1)[/tex]

The given expression is a second  degree polynomial expression with one variable and  roots for this expression can be found out by putting y =0

Hence on putting y=0 in equation (1) we get

[tex]\rm 2x^2 +x -6 = 0[/tex]

On solving for the roots of the equation

For a quadratic equation [tex]\rm ax^2 + bx +c =0 \\According \; to\; Shridharacharya's \; formula\; the\; roots \; of \; the\; equation\; are \; given\; by \\(-b +\sqrt{(b^2 -4ac))}/(2a)} \; and \; (-b -\sqrt{(b^2 -4ac))}/(2a)}............(2)[/tex]  

So on finding the roots [tex]\alpha[/tex] and  [tex]\beta[/tex] from equation (2) we get

[tex]\rm \alpha = \dfrac{-1+\sqrt{1^2 -4\times (2)\times(-6)} }{2\times 2} = 6/4 = 3/2 =1.5[/tex]

[tex]\rm \beta = \dfrac{-1-\sqrt{1^2 -4\times (2)\times(-6)} }{2\times 2} = -8/4 = -2[/tex]

So 2 zero pairs needed to model this polynomial .

The equivalent factored form is

[tex]\rm\bold{(x-3/2)(x+2) =0}[/tex]

For more information please refer to the link below

https://brainly.com/question/23033812

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