Solve 2x ^ 2 - 8x + 8 and (x - 2) ^ 2 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$2 x ^ { 2 } - 8 x + 8 ( x - 2 ) ^ { 2 }$$

Evaluate

$10x^{2}-40x+32$

Solution Steps

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-2\right)^{2}$.

$$2x^{2}-8x+8\left(x^{2}-4x+4\right)$$

Use the distributive property to multiply $8$ by $x^{2}-4x+4$.

$$2x^{2}-8x+8x^{2}-32x+32$$

Combine $2x^{2}$ and $8x^{2}$ to get $10x^{2}$.

$$10x^{2}-8x-32x+32$$

Combine $-8x$ and $-32x$ to get $-40x$.

$$10x^{2}-40x+32$$

Expand

$10x^{2}-40x+32$

Solution Steps

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-2\right)^{2}$.

$$2x^{2}-8x+8\left(x^{2}-4x+4\right)$$

Use the distributive property to multiply $8$ by $x^{2}-4x+4$.

$$2x^{2}-8x+8x^{2}-32x+32$$

Combine $2x^{2}$ and $8x^{2}$ to get $10x^{2}$.

$$10x^{2}-8x-32x+32$$

Combine $-8x$ and $-32x$ to get $-40x$.

$$10x^{2}-40x+32$$