Objective: Add, subtract, 1. Which polynomial is twice the sum of \( 4 x^{2}-x+1 \) and \( -6 x^{2}+x-4 ? \) a. \( -2 x^{2}-3 \) b. \( -4 x^{2}-3 \) c. \( -4 x^{2}-6 \) d. \( -2 x^{2}+x-5 \) 2. Which expression is equivalent to \( 2\left(x^{2}-1\right)+3 x(x-4) ? \) a. \( 5 x^{2}-5 \) b. \( 5 x^{2}-6 \) c. \( 5 x^{2}-12 x-1 \) d. \( 5 x^{2}-12 x-2 \) \[ \left(3 x^{2}+4 x-8\right)-\left(-2 x^{2}+4 x+2\right) \] 7. Subtract \( 5 x^{2}+2 x-11 \) from \( 3 x^{2}+8 x-7 \). 8. Express the product of \( 2 x^{2}+7 x-10 \) and \( x+5 \) in standard form. Subtract \( 5 x^{2}+2 x-11 \) as trinomial. Express the result as Express the result as a trino which polynomial is equivalent to \( 2(y+z) \) a. \( 6 x^{3}+4 x^{2}-34 \) b. \( 6 x^{3}+3 x^{2}-17 \) c. \( 6 x^{3}+3 x^{2}-22 \) d. \( 6 x^{3}+2 x^{2}-17 \) 1
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The Deep Dive
To find the polynomial that is twice the sum of \( 4 x^{2} - x + 1 \) and \( -6 x^{2} + x - 4 \), first, we sum the two polynomials: \[ (4 x^{2} - x + 1) + (-6 x^{2} + x - 4) = -2 x^{2} - 3. \] Now, doubling this result gives us \( -4 x^{2} - 6 \). So, the answer is choice **c**: \( -4 x^{2} - 6 \). Now, for the second expression, let's simplify \( 2(x^{2} - 1) + 3x(x - 4) \): \[ = 2x^{2} - 2 + 3x^{2} - 12x = 5x^{2} - 12x - 2. \] Thus, the answer is choice **d**: \( 5 x^{2} - 12 x - 2 \). When subtracting \( 5 x^{2} + 2 x - 11 \) from \( 3 x^{2} + 8 x - 7 \): \[ (3 x^{2} + 8 x - 7) - (5 x^{2} + 2 x - 11) = 3 x^{2} + 8 x - 7 - 5 x^{2} - 2 x + 11 = -2 x^{2} + 6 x + 4. \] Expressed as a trinomial: \[ -2x^{2} + 6x + 4. \] Lastly, to express the product of \( 2x^{2} + 7x - 10 \) and \( x + 5 \): \[ (2x^{2} + 7x - 10)(x + 5) = 2x^{3} + 10x^{2} + 7x^{2} + 35x - 10x - 50 = 2x^{3} + 17x^{2} + 25x - 50, \] which is in standard form. To express \( 2(y + z) \) gives you \( 2y + 2z \), which looks more manageable when dealing with algebraic equations!
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