Question

help me

Nonsense/No solution = report​

Answer 1

$\large\underline{\sf{Solution-1}}$

$\rm \: {x}^{2} + 8x + 7$

$\rm \: = \: {x}^{2} + 7x + x + 7$

$\rm \: = \: x(x + 7) + 1(x + 7)$

$\rm \: = \: (x + 7)(x + 1)$

Hence,

$\rm\implies \:\boxed{\tt{ {x}^{2} + 8x + 7 = \: (x + 7)(x + 1) \: }}$

$\red{\large\underline{\sf{Solution-2}}}$

$\rm \: {x}^{2} - 11x + 10$

$\rm \: = \: {x}^{2} - 10x - x + 10$

$\rm \: = \: x(x - 10) - 1(x - 10)$

$\rm \: = \: (x - 10)(x - 1)$

Hence,

$\rm\implies \:\boxed{\tt{ {x}^{2} - 11x + 10 = \: (x - 10)(x - 1) \: }}$

$\red{\large\underline{\sf{Solution-3}}}$

$\rm \: {x}^{2} + x - 90$

$\rm \: = \: {x}^{2} + 10x - 9x - 90$

$\rm \: = \: x(x + 10) - 9(x + 10)$

$\rm \: = \: (x + 10)(x - 9)$

Hence,

$\rm\implies \:\boxed{\tt{ {x}^{2} + x - 90 = (x + 10)(x - 9) \: }}$

$\red{\large\underline{\sf{Solution-4}}}$

$\rm \: {x}^{2} + 4x - 12$

$\rm \: = \: {x}^{2} + 6x - 2x - 12$

$\rm \: = \: x(x + 6) - 2(x + 6)$

$\rm \: = \: (x + 6)(x - 2)$

Hence,

$\rm\implies \:\boxed{\tt{ {x}^{2} + 4x - 12 = \: (x + 6)(x - 2) \: }}$

$\red{\large\underline{\sf{Solution-5}}}$

$\rm \: {x}^{2} - 10x + 9$

$\rm \: = \: {x}^{2} - 9x - x+ 9$

$\rm \: = \: x(x - 9) - 1(x - 9)$

$\rm \: = \: (x - 9)(x - 1)$

Hence,

$\rm\implies \:\boxed{\tt{ {x}^{2} - 10x + 9 = \: (x - 9)(x - 1) \: }}$

$\red{\large\underline{\sf{Solution-6}}}$

$\rm \: {3x}^{2} - 2x - 5$

$\rm \: = \: {3x}^{2} - 5x + 3x - 5$

$\rm \: = \: x(3x - 5) + 1(3x - 5)$

$\rm \: = \: (3x - 5)(x + 1)$

Hence,

$\rm\implies \:\boxed{\tt{ {3x}^{2} - 2x - 5 = \: (3x - 5)(x + 1) \: }}$

$\red{\large\underline{\sf{Solution-7}}}$

$\rm \: {2x}^{2} + 3x - 9$

$\rm \: = \: {2x}^{2} + 6x - 3x - 9$

$\rm \: = \: 2x(x + 3) - 3(x + 3)$

$\rm \: = \: (x + 3)(2x - 3)$

Hence,

$\rm\implies \:\boxed{\tt{ {2x}^{2} + 3x - 9 = \: (x + 3)(2x - 3) \: }}$

So,

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Quadratic \: expression & \bf Factors \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf {x}^{2} + 8x + 7 & \sf (x + 1)(x + 7) \\ \\ \sf {x}^{2} - 11x + 10 & \sf (x - 10)(x - 1)\\ \\ \sf {x}^{2} + x - 90 & \sf (x + 10)(x - 9)\\ \\ \sf {x}^{2} + 4x - 12 & \sf (x + 6)(x - 2)\\ \\ \sf {x}^{2} - 10 + 9 & \sf (x - 9)(x - 1)\\ \\ \sf {3x}^{2} - 2x - 5 & \sf (x + 1)(3x - 5) \\ \\ \sf 2 {x}^{2} + 3x - 9 & \sf (x + 3)(2x - 3) \end{array}} \\ \end{gathered}$

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BASIC CONCEPT USED

Splitting of middle terms :-

In order to factorize  ax² + bx + c we have to split b in terms of m and n in such a way that m + n = b and mn = ac.

After finding m and n, we split the middle term b in the quadratic expression as mx + nx and get required factors by grouping the terms.

Answer 2

$\large\underline{\sf{Solution-1}}$

$\rm \: {x}^{2} + 8x + 7$

$\rm \: = \: {x}^{2} + 7x + x + 7$

$\rm \: = \: x(x + 7) + 1(x + 7)$

$\rm \: = \: (x + 7)(x + 1)$

Hence,

$\rm\implies \:\boxed{\tt{ {x}^{2} + 8x + 7 = \: (x + 7)(x + 1) \: }}$

$\red{\large\underline{\sf{Solution-2}}}$

$\rm \: {x}^{2} - 11x + 10$

$\rm \: = \: {x}^{2} - 10x - x + 10$

$\rm \: = \: x(x - 10) - 1(x - 10)$

$\rm \: = \: (x - 10)(x - 1)$

Hence,

$\rm\implies \:\boxed{\tt{ {x}^{2} - 11x + 10 = \: (x - 10)(x - 1) \: }}$

$\red{\large\underline{\sf{Solution-3}}}$

$\rm \: {x}^{2} + x - 90$

$\rm \: = \: {x}^{2} + 10x - 9x - 90$

$\rm \: = \: x(x + 10) - 9(x + 10)$

$\rm \: = \: (x + 10)(x - 9)$

Hence,

$\rm\implies \:\boxed{\tt{ {x}^{2} + x - 90 = (x + 10)(x - 9) \: }}$

$\red{\large\underline{\sf{Solution-4}}}$

$\rm \: {x}^{2} + 4x - 12$

$\rm \: = \: {x}^{2} + 6x - 2x - 12$

$\rm \: = \: x(x + 6) - 2(x + 6)$

$\rm \: = \: (x + 6)(x - 2)$

Hence,

$\rm\implies \:\boxed{\tt{ {x}^{2} + 4x - 12 = \: (x + 6)(x - 2) \: }}$

$\red{\large\underline{\sf{Solution-5}}}$

$\rm \: {x}^{2} - 10x + 9$

$\rm \: = \: {x}^{2} - 9x - x+ 9$

$\rm \: = \: x(x - 9) - 1(x - 9)$

$\rm \: = \: (x - 9)(x - 1)$

Hence,

$\rm\implies \:\boxed{\tt{ {x}^{2} - 10x + 9 = \: (x - 9)(x - 1) \: }}$

$\red{\large\underline{\sf{Solution-6}}}$

$\rm \: {3x}^{2} - 2x - 5$

$\rm \: = \: {3x}^{2} - 5x + 3x - 5$

$\rm \: = \: x(3x - 5) + 1(3x - 5)$

$\rm \: = \: (3x - 5)(x + 1)$

Hence,

$\rm\implies \:\boxed{\tt{ {3x}^{2} - 2x - 5 = \: (3x - 5)(x + 1) \: }}$

$\red{\large\underline{\sf{Solution-7}}}$

$\rm \: {2x}^{2} + 3x - 9$

$\rm \: = \: {2x}^{2} + 6x - 3x - 9$

$\rm \: = \: 2x(x + 3) - 3(x + 3)$

$\rm \: = \: (x + 3)(2x - 3)$

Hence,

$\rm\implies \:\boxed{\tt{ {2x}^{2} + 3x - 9 = \: (x + 3)(2x - 3) \: }}$

So,

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Quadratic \: expression & \bf Factors \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf {x}^{2} + 8x + 7 & \sf (x + 1)(x + 7) \\ \\ \sf {x}^{2} - 11x + 10 & \sf (x - 10)(x - 1)\\ \\ \sf {x}^{2} + x - 90 & \sf (x + 10)(x - 9)\\ \\ \sf {x}^{2} + 4x - 12 & \sf (x + 6)(x - 2)\\ \\ \sf {x}^{2} - 10 + 9 & \sf (x - 9)(x - 1)\\ \\ \sf {3x}^{2} - 2x - 5 & \sf (x + 1)(3x - 5) \\ \\ \sf 2 {x}^{2} + 3x - 9 & \sf (x + 3)(2x - 3) \end{array}} \\ \end{gathered}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

BASIC CONCEPT USED

Splitting of middle terms :-

In order to factorize  ax² + bx + c we have to split b in terms of m and n in such a way that m + n = b and mn = ac.

After finding m and n, we split the middle term b in the quadratic expression as mx + nx and get required factors by grouping the terms.