Question

type the correct answer in each box. use numerals instead of words.
simplify the following polynomial expression.
$(5x^2 + 13x - 4) - (17x^2 + 7x - 19) + (5x - 7)(3x + 1)$
$square x^2 - square x + square$
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Explanation:

Step1: Expand and simplify term by term

First, handle the subtraction: \((5x^2 + 13x - 4) - (17x^2 + 7x - 19)=5x^2 + 13x - 4 - 17x^2 - 7x + 19\)
Combine like terms: \((5x^2 - 17x^2)+(13x - 7x)+(- 4 + 19)= - 12x^2 + 6x + 15\)

Next, expand \((5x - 7)(3x + 1)\):
Using the distributive property (FOIL method): \(5x\times3x + 5x\times1 - 7\times3x - 7\times1 = 15x^2 + 5x - 21x - 7 = 15x^2 - 16x - 7\)

Now, add the two simplified expressions together: \((- 12x^2 + 6x + 15)+(15x^2 - 16x - 7)\)
Combine like terms: \((-12x^2 + 15x^2)+(6x - 16x)+(15 - 7)=3x^2 - 10x + 8\)

Wait, but the problem seems to have a typo or maybe I misread. Wait, the original expression is \((5x^2 + 13x - 4)-(17x^2 + 7x - 19)+(5x - 7)(3x + 1)\)? Wait, no, looking back, maybe the user's problem is to simplify \((5x^2 + 13x - 4)-(17x^2 + 7x - 19)+(5x - 7)(3x + 1)\) and then express as \(\square x^2-\square x+\square\)? Wait, no, the boxes are for \(x^2 - \square x + \square\)? Wait, no, the left side is \(x^2 - \square x + \square\)? Wait, maybe I made a mistake. Wait, let's re - do the expansion:

Wait, maybe the problem is \((5x^2 + 13x - 4)-(17x^2 + 7x - 19)+(5x - 7)(3x + 1)\) and then we need to write it in the form \(ax^2+bx + c\)

First, \((5x^2 + 13x - 4)-(17x^2 + 7x - 19)=5x^2+13x - 4 - 17x^2 - 7x + 19=-12x^2 + 6x + 15\)

\((5x - 7)(3x + 1)=15x^2+5x - 21x - 7 = 15x^2-16x - 7\)

Now add them: \(-12x^2 + 6x + 15+15x^2-16x - 7=( - 12x^2+15x^2)+(6x - 16x)+(15 - 7)=3x^2-10x + 8\)

But the left side of the boxes is \(x^2-\square x+\square\)? Wait, that doesn't match. Wait, maybe the original problem is \((5x^2 + 13x - 4)-(17x^2 + 7x - 19)+(5x - 7)(3x + 1)\) and we need to write it as \(3x^2-10x + 8\), but if the first term is \(x^2\), maybe there is a miscalculation. Wait, let's check the multiplication again:

\((5x - 7)(3x + 1)=5x\times3x+5x\times1-7\times3x - 7\times1 = 15x^2+5x - 21x - 7=15x^2-16x - 7\)

\((5x^2 + 13x - 4)-(17x^2 + 7x - 19)=5x^2-17x^2+13x - 7x-4 + 19=-12x^2+6x + 15\)

Adding \(-12x^2+6x + 15\) and \(15x^2-16x - 7\):

\((-12x^2+15x^2)=3x^2\), \((6x-16x)=-10x\), \((15 - 7)=8\). So the result is \(3x^2-10x + 8\)

But the problem's left - hand side of the boxes is \(x^2-\square x+\square\), which is different. Maybe the original problem was \((5x^2 + 13x - 4)-(17x^2 + 7x - 19)+(5x - 7)(3x + 1)\) and we need to factor or there is a mistake. Wait, maybe the user intended a different operation. Alternatively, maybe the first part is \((5x^2 + 13x - 4)-(17x^2 + 7x - 19)\) and then add \((5x - 7)(3x + 1)\), but the way the boxes are set up is \(x^2-\square x+\square\), which would mean the coefficient of \(x^2\) is 1, so maybe I made a mistake in the sign. Wait, let's re - check the subtraction:

\((5x^2 + 13x - 4)-(17x^2 + 7x - 19)=5x^2-17x^2+13x - 7x-4 + 19=-12x^2+6x + 15\)

\((5x - 7)(3x + 1)=15x^2-16x - 7\)

Adding them: \(-12x^2+15x^2=3x^2\), \(6x-16x=-10x\), \(15 - 7 = 8\). So \(3x^2-10x + 8\). If we factor this quadratic: \(3x^2-10x + 8=(3x - 4)(x - 2)\), but that's not helpful. Wait, maybe the original problem was \((5x^2 + 13x - 4)-(17x^2 + 7x - 19)+(5x - 7)(3x + 1)\) and the boxes are for the coefficients when written as \(ax^2+bx + c\), so \(a = 3\), \(b=-10\), \(c = 8\)

But the way the problem is presented with \(x^2-\square x+\square\) suggests that maybe there is a mistake in my calculation. Wait, let's check the multiplication of \((5x - 7)(3x + 1)\) again:

\(5x\times3x = 15x^2\), \(5x\times1=5x\), \(-7\times3x=-21x\), \(-7\times1=-7\). So \(15x^2+5x-21x - 7=15x^2-16x - 7\). Correct.

Subtraction: \(5x^2-17x^2=-12x^2\), \(13x-7x = 6x\), \(-4 + 19=15\). Correct.

Adding: \(-12x^2+15x^2=3x^2\), \(6x-16x=-10x\), \(15 - 7 = 8\). So the result is \(3x^2-10x + 8\)

If we divide the entire expression by 3, we get \(x^2-\frac{10}{3}x+\frac{8}{3}\), but that's not an integer. So maybe the original problem has a different set of parentheses. Alternatively, maybe the expression is \((5x^2 + 13x - 4)-(17x^2 + 7x - 19)+(5x - 7)(3x - 1)\) (changing the last sign). Let's try that:

\((5x - 7)(3x - 1)=15x^2-5x-21x + 7=15x^2-26x + 7\)

Then adding to \(-12x^2+6x + 15\): \(-12x^2+15x^2+6x-26x + 15 + 7=3x^2-20x + 22\), still not helpful.

Alternatively, maybe the first subtraction is \((17x^2 + 7x - 19)-(5x^2 + 13x - 4)\):

\(17x^2-5x^2+7x-13x-19 + 4=12x^2-6x - 15\)

Then adding \((5x - 7)(3x + 1)=15x^2-16x - 7\):

\(12x^2+15x^2-6x-16x-15 - 7=27x^2-22x - 22\), no.

Given that, I think the correct simplification is \(3x^2-10x + 8\), so if we consider the form \(ax^2+bx + c\), \(a = 3\), \(b=-10\), \(c = 8\)

But the problem's boxed form is \(x^2-\square x+\square\), which is confusing. Maybe there was a typo and the original expression was different. However, based on the given expression, the simplified form is \(3x^2-10x + 8\)

Wait, maybe the user made a mistake in the problem statement. Alternatively, maybe I misread the original expression. Let's re - examine the image:

The expression is:

Simplify the following polynomial expression:

\((5x^2 + 13x - 4)-(17x^2 + 7x - 19)+(5x - 7)(3x + 1)\)

And then write it as \(\square x^2-\square x+\square\)

So the coefficients are 3, 10, 8 (since it's \(3x^2-10x + 8\))

Answer:

3, 10, 8 (in the order of the coefficients for \(x^2\), the coefficient of \(x\) (absolute value, since it's \(-10x\) so the box for \(x\) is 10), and the constant term 8)