Question

what is the value of the missing term?
$(square + 4)(x^2 - 3x + 2) = -5x^3 + 19x^2 - 22x + 8$

a. $5$
b. $5x$
c. $-5x$
d. $-5$

Explanation:

Step1: Let the missing term be \( ax \) (wait, actually, let's assume the missing term is \( mx \)? No, wait, the left side is \( (\square + 4)(x^2 - 3x + 2) \), and the right side has \( -5x^3 \), so the missing term must be a linear term in \( x \) (or maybe a constant? Wait, no, when we multiply \( \square \) (let's call it \( A \)) with \( x^2 \), we get \( A x^2 \times x \)? Wait, no, wait: \( (A + 4)(x^2 - 3x + 2) \). Let's expand the left side:

\( A(x^2 - 3x + 2) + 4(x^2 - 3x + 2) \)

\( = A x^2 - 3A x + 2A + 4x^2 - 12x + 8 \)

Now, combine like terms:

For \( x^3 \)? Wait, wait, no! Wait, the right side has \( -5x^3 \), but the left side, if \( A \) is a linear term (like \( bx \)), then \( bx \times x^2 = bx^3 \). Oh! I made a mistake. The missing term is a linear term? No, wait, the left side is \( (\square + 4)(x^2 - 3x + 2) \). So \( \square \) must be a linear term (degree 1) because when multiplied by \( x^2 \) (degree 2), we get degree 3, which matches the right side's \( -5x^3 \). So let \( \square = mx \). Then:

\( (mx + 4)(x^2 - 3x + 2) = mx(x^2 - 3x + 2) + 4(x^2 - 3x + 2) \)

\( = mx^3 - 3mx^2 + 2mx + 4x^2 - 12x + 8 \)

Now, combine like terms:

- \( x^3 \) term: \( m x^3 \)
- \( x^2 \) term: \( (-3m + 4) x^2 \)
- \( x \) term: \( (2m - 12) x \)
- constant term: \( 8 \)

Now, set equal to the right side: \( -5x^3 + 19x^2 - 22x + 8 \)

So equate coefficients:

1. \( x^3 \) term: \( m = -5 \)? Wait, no, wait: \( m x^3 = -5x^3 \) ⇒ \( m = -5 \)? Wait, but let's check the \( x^2 \) term:

\( -3m + 4 = 19 \)

If \( m = -5 \), then \( -3(-5) + 4 = 15 + 4 = 19 \), which matches the \( x^2 \) coefficient (19). Let's check the \( x \) term:

\( 2m - 12 = 2(-5) - 12 = -10 - 12 = -22 \), which matches the \( x \) coefficient (-22). And the constant term is 8, which matches. So the missing term is \( -5x \)? Wait, no, wait: \( \square = mx \), and \( m = -5 \), so \( \square = -5x \)? Wait, but let's check the options. Option c is \( -5x \). Wait, but let's re-express:

Wait, when we set \( \square = A \), and when we expand, the \( x^3 \) term comes from \( A \times x^2 \), so \( A \) must be a linear term (degree 1) to get \( x^3 \) (degree 3). So \( A = -5x \), because \( -5x \times x^2 = -5x^3 \), which matches the right side's \( -5x^3 \). Then, let's verify:

\( (-5x + 4)(x^2 - 3x + 2) \)

Multiply \( -5x \) by \( x^2 - 3x + 2 \): \( -5x^3 + 15x^2 - 10x \)

Multiply 4 by \( x^2 - 3x + 2 \): \( 4x^2 - 12x + 8 \)

Now, add them together:

\( -5x^3 + 15x^2 - 10x + 4x^2 - 12x + 8 \)

Combine like terms:

\( -5x^3 + (15x^2 + 4x^2) + (-10x - 12x) + 8 \)

\( = -5x^3 + 19x^2 - 22x + 8 \), which matches the right side. So the missing term is \( -5x \), which is option c.

Wait, but let's check the steps again. Let's denote the missing term as \( A \). Then:

\( (A + 4)(x^2 - 3x + 2) = -5x^3 + 19x^2 - 22x + 8 \)

Expand left side:

\( A(x^2 - 3x + 2) + 4(x^2 - 3x + 2) \)

\( = A x^2 - 3A x + 2A + 4x^2 - 12x + 8 \)

Wait, but this would give \( x^2 \) terms, not \( x^3 \). Oh! Wait, I made a mistake earlier. The left side is \( (A + 4)(x^2 - 3x + 2) \), but the right side has \( x^3 \), so \( A \) must be a linear term (degree 1) to get \( x^3 \) when multiplied by \( x^2 \) (degree 2). So \( A \) is a linear term, say \( A = mx \). Then:

\( (mx + 4)(x^2 - 3x + 2) = mx(x^2 - 3x + 2) + 4(x^2 - 3x + 2) \)

\( = mx^3 - 3mx^2 + 2mx + 4x^2 - 12x + 8 \)

Now, combine like terms:

- \( x^3 \): \( m x^3 \)
- \( x^2 \): \( (-3m + 4) x^2 \)
- \( x \): \( (2m - 12) x \)
- constant: 8

Now, set equal to right side: \( -5x^3 + 19x^2 - 22x + 8 \)

So:

1. \( x^3 \) coefficient: \( m = -5 \) (since \( m x^3 = -5x^3 \))
2. \( x^2 \) coefficient: \( -3m + 4 = 19 \). Substitute \( m = -5 \): \( -3(-5) + 4 = 15 + 4 = 19 \), which matches.
3. \( x \) coefficient: \( 2m - 12 = 2(-5) - 12 = -10 - 12 = -22 \), which matches.
4. Constant term: 8, which matches.

Thus, the missing term \( A = mx = -5x \), which is option c.

Step2: Verify by substituting \( -5x \) into the left side:

\( (-5x + 4)(x^2 - 3x + 2) \)

First, multiply \( -5x \) by each term in \( x^2 - 3x + 2 \):

\( -5x \times x^2 = -5x^3 \)

\( -5x \times (-3x) = 15x^2 \)

\( -5x \times 2 = -10x \)

Then, multiply 4 by each term in \( x^2 - 3x + 2 \):

\( 4 \times x^2 = 4x^2 \)

\( 4 \times (-3x) = -12x \)

\( 4 \times 2 = 8 \)

Now, add all these terms together:

\( -5x^3 + 15x^2 - 10x + 4x^2 - 12x + 8 \)

Combine like terms:

- \( x^3 \): \( -5x^3 \)
- \( x^2 \): \( 15x^2 + 4x^2 = 19x^2 \)
- \( x \): \( -10x - 12x = -22x \)
- constant: \( 8 \)

Which gives \( -5x^3 + 19x^2 - 22x + 8 \), matching the right side. So the missing term is \( -5x \).

Answer:

c. -5x