Question

q: factor the polynomial: drag and drop the factors of the polynomial to complete the equation.

16x⁴ - 8x² + 1 =

factors: 4x, 8x, 16x, (4x + 1), (4x - 1), (2x + 1), (2x - 1), (2x + 2), (2x - 2), (x - 2), (x + 2), (16x - 1), (16x + 1)

Explanation:

Step1: Let \( y = x^2 \), rewrite the polynomial.

The polynomial \( 16x^4 - 8x^2 + 1 \) can be rewritten as \( 16y^2 - 8y + 1 \) by substituting \( y = x^2 \).

Step2: Factor the quadratic in \( y \).

We recognize \( 16y^2 - 8y + 1 \) as a perfect square trinomial. The formula for a perfect square trinomial is \( a^2 - 2ab + b^2=(a - b)^2 \). Here, \( a = 4y \) and \( b = 1 \), so \( 16y^2 - 8y + 1=(4y - 1)^2 \).

Step3: Substitute back \( y = x^2 \).

Substituting \( y = x^2 \) back in, we get \( (4x^2 - 1)^2 \).

Step4: Factor the difference of squares.

We know that \( 4x^2 - 1 \) is a difference of squares, which factors as \( (2x + 1)(2x - 1) \). So \( (4x^2 - 1)^2 = [(2x + 1)(2x - 1)]^2=(2x + 1)^2(2x - 1)^2 \).
Looking at the given factors, we can also factor \( 16x^4 - 8x^2 + 1 \) as \( (4x^2 - 1)^2=(4x + 1)(4x - 1)(2x + 1)(2x - 1) \)? Wait, no, actually \( 4x^2 - 1=(2x + 1)(2x - 1) \), so squaring that gives \( (2x + 1)^2(2x - 1)^2 \). But from the options, let's check:
Wait, let's do it step by step again. Let's consider the original polynomial \( 16x^4 - 8x^2 + 1 \). Let's try to factor it as a quadratic in \( x^2 \). Let \( u = x^2 \), then the polynomial is \( 16u^2 - 8u + 1 \). The discriminant of \( 16u^2 - 8u + 1 \) is \( (-8)^2 - 4\times16\times1 = 64 - 64 = 0 \), so it's a perfect square. \( 16u^2 - 8u + 1=(4u - 1)^2 \). Substituting back \( u = x^2 \), we get \( (4x^2 - 1)^2 \). Now, \( 4x^2 - 1 \) is a difference of squares: \( 4x^2 - 1=(2x + 1)(2x - 1) \). So \( (4x^2 - 1)^2=(2x + 1)^2(2x - 1)^2 \). But looking at the options, we have factors like \( (4x + 1) \), \( (4x - 1) \), etc. Wait, maybe I made a mistake. Let's try another approach. Let's suppose \( 16x^4 - 8x^2 + 1=(ax^2 + bx + c)(dx^2 + ex + f) \). Since there are no \( x^3 \) or \( x \) terms, \( b = e = 0 \). So it's \( (ax^2 + c)(dx^2 + f)=adx^4+(af + cd)x^2 + cf \). So \( ad = 16 \), \( af + cd=-8 \), \( cf = 1 \). Let's take \( a = 4 \), \( d = 4 \), \( c = -1 \), \( f = -1 \). Then \( af + cd=4\times(-1)+4\times(-1)=-8 \), which works. So \( (4x^2 - 1)(4x^2 - 1)=(4x^2 - 1)^2 \), which is the same as before. Now, \( 4x^2 - 1=(2x + 1)(2x - 1) \), so \( (4x^2 - 1)^2=(2x + 1)^2(2x - 1)^2 \). But the options include \( (4x + 1) \), \( (4x - 1) \), etc. Wait, maybe the polynomial can be factored as \( (4x^2 - 1)^2=(4x + 1)(4x - 1)(2x + 1)(2x - 1) \)? No, that's not correct because \( (4x + 1)(4x - 1)=16x^2 - 1 \), and \( (2x + 1)(2x - 1)=4x^2 - 1 \), so \( (16x^2 - 1)(4x^2 - 1)=64x^4 - 20x^2 + 1 \), which is not the original polynomial. So my initial approach is correct. Wait, the original polynomial is \( 16x^4 - 8x^2 + 1 \). Let's check \( (4x^2 - 1)^2=(4x^2)^2 - 2\times4x^2\times1 + 1^2=16x^4 - 8x^2 + 1 \), which is correct. So the factors are \( (4x^2 - 1)^2 \), and \( 4x^2 - 1=(2x + 1)(2x - 1) \), so the factors are \( (2x + 1) \), \( (2x - 1) \), \( (2x + 1) \), \( (2x - 1) \). But looking at the options, we have \( (4x + 1) \), \( (4x - 1) \), etc. Wait, maybe there's a mistake in my factoring. Let's try to factor \( 16x^4 - 8x^2 + 1 \) as a quadratic in \( x^2 \) with different coefficients. Let \( a = 16 \), \( b = -8 \), \( c = 1 \). The roots of \( 16u^2 - 8u + 1 = 0 \) are \( u=\frac{8\pm\sqrt{64 - 64}}{32}=\frac{8}{32}=\frac{1}{4} \). So \( 16u^2 - 8u + 1=16(u - \frac{1}{4})^2=(4u - 1)^2 \), which is the same as before. So \( 16x^4 - 8x^2 + 1=(4x^2 - 1)^2=(2x + 1)^2(2x - 1)^2 \). So the factors are \( (2x + 1) \), \( (2x - 1) \), \( (2x + 1) \), \( (2x - 1) \). But the options include \( (4x + 1) \), \( (4x - 1) \), etc. Wait, maybe the problem is written as \( 16x^4 - 8x^2 + 1 \), and the options are supposed to be dragged to form the factors. Let's check the product of \( (4x^2 - 1)^2 \), but the options have linear factors? Wait, no, the options have factors like \( 4x \), \( 8x \), \( 16x \), and linear factors in \( x \). Wait, maybe I misread the polynomial. Is it \( 16x^4 - 8x^3 + 1 \)? No, the user wrote \( 16x^4 - 8x^2 + 1 \). Wait, maybe the polynomial is \( 16x^4 - 8x^2 + 1 \), and we can factor it as \( (4x^2 - 1)^2=(4x + 1)(4x - 1)(2x + 1)(2x - 1) \)? No, that's incorrect. Wait, \( (4x + 1)(4x - 1)=16x^2 - 1 \), and \( (2x + 1)(2x - 1)=4x^2 - 1 \), so multiplying those two gives \( (16x^2 - 1)(4x^2 - 1)=64x^4 - 20x^2 + 1 \), which is not the original polynomial. So there must be a mistake in my approach. Let's try to factor \( 16x^4 - 8x^2 + 1 \) as a quadratic in \( x^2 \) with \( a = 4 \), \( d = 4 \), \( c = -1 \), \( f = -1 \), so \( (4x^2 - 1)(4x^2 - 1) \), which is correct. Now, \( 4x^2 - 1=(2x + 1)(2x - 1) \), so the factors are \( (2x + 1) \), \( (2x - 1) \), \( (2x + 1) \), \( (2x - 1) \). But the options include \( (4x + 1) \), \( (4x - 1) \), etc. Wait, maybe the problem is \( 16x^4 - 8x^3 + 1 \), but no, the user wrote \( 16x^4 - 8x^2 + 1 \). Alternatively, maybe the polynomial is \( 16x^4 - 8x^2 + 1 \), and the correct factors are \( (4x^2 - 1)^2 \), but the options have linear factors, so maybe the problem is actually \( 16x^4 - 8x^3 + x^2 + \dots \)? No, the user's image shows \( 16x^4 - 8x^2 + 1 \). Wait, let's check the product of \( (4x^2 - 1)^2 \), which is \( 16x^4 - 8x^2 + 1 \), so that's correct. Now, if we consider that the options are linear factors, maybe the problem is miswritten, or I'm missing something. Wait, maybe the polynomial is \( 16x^4 - 8x^3 + 1 \), but no. Alternatively, maybe the factors are \( (4x^2 - 1)^2=(4x + 1)(4x - 1)(2x + 1)(2x - 1) \), but that's not correct. Wait, no, \( (4x + 1)(4x - 1)=16x^2 - 1 \), and \( (2x + 1)(2x - 1)=4x^2 - 1 \), so \( (16x^2 - 1)(4x^2 - 1)=64x^4 - 20x^2 + 1 \), which is not the original polynomial. So I think the correct factors are \( (2x + 1) \), \( (2x - 1) \), \( (2x + 1) \), \( (2x - 1) \), so we need to drag \( (2x + 1) \), \( (2x - 1) \), \( (2x + 1) \), \( (2x - 1) \) to the box. But the options include \( (4x + 1) \), \( (4x - 1) \), etc. Wait, maybe the polynomial is \( 16x^4 - 8x^2 + 1 \), and the factors are \( (4x^2 - 1)^2 \), but the options have \( (4x + 1) \) and \( (4x - 1) \) as factors of \( 4x^2 - 1 \)? No, \( 4x^2 - 1=(2x + 1)(2x - 1) \), not \( (4x + 1)(4x - 1) \). Wait, \( (4x + 1)(4x - 1)=16x^2 - 1 \), which is a different quadratic. So I'm confused. Maybe the problem is \( 16x^4 - 8x^3 + x^2 + \dots \), but the user wrote \( 16x^4 - 8x^2 + 1 \). Alternatively, maybe the correct factors are \( (4x^2 - 1)^2 \), and the options have \( (4x + 1) \) and \( (4x - 1) \) as mistakes, but the intended factors are \( (2x + 1) \) and \( (2x - 1) \) squared. So the answer is to drag \( (2x + 1) \), \( (2x - 1) \), \( (2x + 1) \), \( (2x - 1) \) to the box. But let's check the product of \( (2x + 1)(2x - 1)(2x + 1)(2x - 1) \). \( (2x + 1)(2x - 1)=4x^2 - 1 \), so squaring that gives \( (4x^2 - 1)^2=16x^4 - 8x^2 + 1 \), which is correct. So the factors are \( (2x + 1) \), \( (2x - 1) \), \( (2x + 1) \), \( (2x - 1) \).

Answer:

The factors are \((2x + 1)\), \((2x - 1)\), \((2x + 1)\), \((2x - 1)\) (dragging these four factors to the box to form \((2x + 1)^2(2x - 1)^2=(4x^2 - 1)^2 = 16x^4 - 8x^2 + 1\)).