Example 1: x^{2}- 12x + 35

Solution:

Get the master product and middle term: Master product = first term * constant = 35x ^{2}

Middle term = -12xSplit the middle term into two parts whose product equals the master product: -12x = -5x - 7x (& -5x * -7x = 35x ^{2}= master product)Rewrite the expression with the middle term split into two parts: x ^{2}- 5x - 7x + 35Make two groups, with each group consisting of two terms: (x ^{2}- 5x) + (-7x + 35)Factor out the common factors from each group: x(x - 5) - 7(x - 5) Take (x - 5) as the common factor: (x - 5)(x - 7) Answer: (x - 5)(x - 7)

Example 2: 8x^{2}- 26x + 15

Solution:

Get the master product and middle term: Master product = first term * constant number = 8x ^{2}* 15 = 120x^{2}

Middle term = -26xSplit the middle term into two parts whose product equals the master product: -26x = -20x - 6x (& -20x * -6x = 120x ^{2})Rewrite the quadratic expression with the two parts: 8x ^{2}- 20x - 6x + 15Group the four termed polynomial so formed into two groups, with each group containing two terms: (8x ^{2}- 20x) + (-6x + 15)Factor out common factors from each group: 4x(2x - 5) - 3(2x - 5) Take (2x - 5) as the common factor: (2x - 5)(4x - 3) Answer: (2x - 5)(4x - 3)

Example 3: 2x^{2}+ 5x - 25

Solution:

Calculate the master product and middle term: Master product = first term * constant = 2x ^{2}* -25 = -50x^{2}

Middle term = 5xDivide the middle term 5x into two parts such that their product equals the master product -50x ^{2}5x = 10x - 5x (& 10x * -5x = -50x ^{2})Rewrite the quadratic expression with the two parts of the middle term: 2x ^{2}+ 10x - 5x - 25Group the four terms into two groups with each group containing two terms: (2x ^{2}+ 10x) + (-5x - 25)Factor out common factors from each group: 2x(x + 5) - 5(x + 5) T(x + 5) is the common factor: (x + 5)(2x - 5) Answer: (x + 5)(2x - 5)

Example 4: √3x^{2}+ 11x + 6√3

Solution:

Calculate the master product and middle term of the given quadratic expression: Master product = first term * constant = √3x ^{2}* 6√3 = 18x^{2}

Middle term = 11xSplit the middle term into two parts such that their product equals the master product: 11x = 9x + 2x (& 9x * 2x = 18x ^{2}= master product of quadratic)Rewrite the quadratic with the two parts of the middle term: √3x ^{2}+ 9x + 2x + 6√3Group into two groups with each group containing two terms: (√3x ^{2}+ 9x) + (2x + 6√3)Take common factors from each group: √3x(x + 3√3) + 2(x + 3√3) Take (x + 3√3) as a common factor: (x + 3√3)(√3x + 3) Answer: (x + 3√3)(√3x + 2)

Example 5: 3a^{2}x^{2}+ 8abx + 4b^{2}

Solution:

Calculate the master product and middle term: Master product = first term * constant(or third term) = 3a ^{2}x^{2}* 4b^{2}= 12a^{2}x^{2}b^{2}

Middle term: 8abxSplit the middle term into two parts whose product equals the master product: 8abx = 6abx + 2abx (& 6abx * 2abx = 12a ^{2}x^{2}b^{2}= master product)Rewrite the quadratic expression with the two parts of the middle term: 3a ^{2}x^{2}+ 6abx + 2abx + 4b^{2}Group the four terms into two groups with each group containing two terms: (3a ^{2}x^{2}+ 6abx) + (2abx + 4b^{2})Take common factors from each group: 3ax(ax + 2b) + 2b(ax + 2b) Take (ax + 2b) as the common factor: (ax + 2b)(3ax + 2b) Answer: (ax + 2b)(3ax + 2b)

I like.

ReplyDeleteI didn't get it after example 1.

Sorta by q. 2

easy by q 3

4&5 look difficult but simple when you try.

Can you solve this:

ReplyDeletex^2 4Root2x 6=0