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R2 plus 10r plus 25

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โˆ™ 2009-11-12 14:27:26

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The quadratic expression: r2+10r+25 = (r+5)(r+5) when factorised

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โˆ™ 2009-11-12 14:27:26
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A polynomial of degree zero is a constant term

The grouping method of factoring can still be used when only some of the terms share a common factor A True B False

The sum or difference of p and q is the of the x-term in the trinomial

A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials

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What is the factor of r2 10r 25 polynomial?

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Equation x2 plus Bx plus c equals 0 has 5 as the sum of its roots and 15 as the sum of the squares of its roots The value of 'c' is equals?

x2 - (-b/a)x + (c/a) = 0 or x2 - (sum of the roots)x + (product of the roots) = 0 Let the roots be r1 and r2. So we have: r1 + r2 = 5 (r1)2 + (r2)2 = 15 r1 = 5 - r2 (express r1 in term of r2) (5 - r2)2 + (r2)2 = 15 25 - 10r2 + (r2)2 + (r2)2 = 15 2(r2)2 - 10r + 25 = 15 (subtract 15 to both sides) 2(r2)2 - 10r + 10 = 0 (divide by 2 to both sides) (r2)2 - 5r + 5 = 0 (use the quadratic formula) r2 = [-b + &- sq root of (b - 4ac)]/2a r2 = {-(-5) + &- sq root of [(-5)2 - 4(1)(5)]}/2(1) = [5 + &- sq root of (25 - 20)]/2 = (5 + &- sq root of 5)/2 r1 = 5 - r2 r1 = 5 - (5 + &- sq root of 5)/2 Thus, when r2 = (5 + sq.root of 5)/2, r1 = (5 - sq.root of 5)/2 or vice versa. Since the given equation is x2 + bx + c = 0, a = 1, then c equals to the product of roots. So that, c = (r1)(r2) = [(5 - sq.root of 5)/2][(5 + sq.root of 5)/2] = [52 - (sq.root of 5)2]/4 = 5


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