QUESTION IMAGE
Question
type the correct answer in each box. use numerals instead of words. simplify the following polynomial expression. \\((5x^2 + 13x - 4) - (17x^2 + 7x - 19) + (5x - 7)(3x + 1)\\) \\(\square\\ x^2 - \square\\ x + \square\\)
Step1: Expand and simplify each part
First, expand \((5x - 7)(3x + 1)\):
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Then, simplify \((5x^2 + 13x - 4) - (17x^2 + 7x - 19)\):
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Step2: Combine all parts
Now, add the result of the expansion and the simplified subtraction:
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Wait, there is a mistake above. Let's re - do the combination:
We have \((5x^{2}+13x - 4)-(17x^{2}+7x - 19)+(15x^{2}-16x - 7)\)
First, handle the first two polynomials:
\(5x^{2}+13x - 4-17x^{2}-7x + 19=(5x^{2}-17x^{2})+(13x - 7x)+(-4 + 19)=-12x^{2}+6x + 15\)
Then add \((15x^{2}-16x - 7)\) to \(-12x^{2}+6x + 15\):
\((-12x^{2}+6x + 15)+(15x^{2}-16x - 7)=(-12x^{2}+15x^{2})+(6x-16x)+(15 - 7)=3x^{2}-10x + 8\)? No, wait:
Wait, the original expression is \((5x^{2}+13x - 4)-(17x^{2}+7x - 19)+(5x - 7)(3x + 1)\)
We found \((5x - 7)(3x + 1)=15x^{2}-16x - 7\)
\((5x^{2}+13x - 4)-(17x^{2}+7x - 19)=5x^{2}+13x - 4-17x^{2}-7x + 19=-12x^{2}+6x + 15\)
Now add \(-12x^{2}+6x + 15\) and \(15x^{2}-16x - 7\):
\(-12x^{2}+6x + 15+15x^{2}-16x - 7=( - 12x^{2}+15x^{2})+(6x-16x)+(15 - 7)=3x^{2}-10x + 8\)? But the form is \(\square x^{2}-\square x+\square\), so we made a mistake in the sign when combining.
Wait, let's start over:
The original expression: \((5x^{2}+13x - 4)-(17x^{2}+7x - 19)+(5x - 7)(3x + 1)\)
First, expand \((5x - 7)(3x + 1)=15x^{2}+5x-21x - 7 = 15x^{2}-16x - 7\)
Then, \((5x^{2}+13x - 4)-(17x^{2}+7x - 19)=5x^{2}+13x - 4-17x^{2}-7x + 19=-12x^{2}+6x + 15\)
Now, add \(-12x^{2}+6x + 15\) and \(15x^{2}-16x - 7\):
\(-12x^{2}+6x + 15+15x^{2}-16x - 7=( - 12x^{2}+15x^{2})+(6x-16x)+(15 - 7)=3x^{2}-10x + 8\)
But the problem's final form is \(\square x^{2}-\square x+\square\), so \(3x^{2}-10x + 8\) can be written as \(3x^{2}-10x + 8\), but let's check the calculation again.
Wait, maybe I made a mistake in the sign when expanding the first subtraction. Let's re - calculate \((5x^{2}+13x - 4)-(17x^{2}+7x - 19)\):
\(5x^{2}+13x - 4-17x^{2}-7x + 19=(5 - 17)x^{2}+(13 - 7)x+(-4 + 19)=-12x^{2}+6x + 15\) (this is correct)
\((5x - 7)(3x + 1)=15x^{2}+5x-21x - 7 = 15x^{2}-16x - 7\) (this is correct)
Now add \(-12x^{2}+6x + 15\) and \(15x^{2}-16x - 7\):
\(x^{2}\) terms: \(-12x^{2}+15x^{2}=3x^{2}\)
\(x\) terms: \(6x-16x=-10x\)
constant terms: \(15 - 7 = 8\)
So the expression is \(3x^{2}-10x + 8\)
But the problem's boxed form is \(\square x^{2}-\square x+\square\), so the coefficients are 3, 10, 8.
Wait, maybe I made a mistake in the sign of the \(x\) term. Let's re - do the entire calculation:
Original expression: \((5x^{2}+13x - 4)-(17x^{2}+7x - 19)+(5x - 7)(3x + 1)\)
First, expand \((5x - 7)(3x + 1)\):
\(5x\times3x=15x^{2}\), \(5x\times1 = 5x\), \(-7\times3x=-21x\), \(-7\times1=-7\)
So \((5x - 7)(3x + 1)=15x^{2}+5x-21x - 7=15x^{2}-16x - 7\)
Then, \((5x^{2}+13x - 4)-(17x^{2}+7x - 19)=5x^{2}+13x - 4-17x^{2}-7x + 19\)
Combine like terms:
\(x^{2}\): \(5x^{2}-17x^{2}=-12x^{2}\)
\(x\): \(13x-7x = 6x\)
constants: \(-4 + 19 = 15\)
So \((5x^{2}+13x - 4)-(17x^{2}+7x - 19)=-12x^{2}+6x + 15\)
Now add \(-12x^{2}+6x + 15\) and \(15x^{2}-16x - 7\):
\(x^{2}\): \(-12x^{2}+15x^{2}=3x^{2}\)
\(x\): \(6x-16x=-10x\)
constants: \(15 - 7 = 8\)…
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3 \(x^{2}\) - 10 \(x\) + 8, so the boxes should be filled with 3, 10, 8 respectively.